- Email: [email protected]
DJKM algebras and non-classical orthogonal polynomials
J. Differential Equations 255 (2013) 2846–2870
Contents lists available at SciVerse ScienceDirect
Journal of Differential Equations www.elsevier.com/locate/jde
DJKM algebras and non-classical orthogonal polynomials Ben Cox a,∗ , Vyacheslav Futorny b , Juan A. Tirao c a b c
Department of Mathematics, College of Charleston, 66 George St., Charleston, SC 29424, USA Department of Mathematics, University of São Paulo, São Paulo, Brazil CIEM-FAMAF, Universidad Nacional de Cordoba, 5000 Cordoba, Argentina
a r t i c l e
i n f o
Article history: Received 27 January 2013 Revised 11 June 2013 Available online 2 August 2013
a b s t r a c t We describe families of polynomials arising in the study of the universal central extensions of Lie algebras introduced by Date, Jimbo, Kashiwara, and Miwa (1983) [6] in their work on the Landau– Lifshitz equations. We show these two families of polynomials satisfy certain fourth order linear differential equations by direct computation and one of the families is a particular collection of associated ultraspherical polynomials. © 2013 Elsevier Inc. All rights reserved.
1. Introduction Date, Jimbo, Kashiwara and Miwa [6] studied integrable systems arising from Landau–Lifshitz differential equation. The hierarchy of this equation is written in terms of free fermions on an elliptic curve. The authors introduced an infinite dimensional Lie algebra which is a one dimensional central extension of g ⊗ C[t , t −1 , u | u 2 = (t 2 − b2 )(t 2 − c 2 )] where b = ±c are complex constants and g is a simple finite dimensional Lie algebra. This Lie algebra which we call the DJKM algebra, acts on the solutions of the Landau–Lifshitz equation as infinitesimal Bäcklund transformations. The Lie algebra above is an example of a Krichever–Novikov algebra (see [15,14,16]). A fair amount of interesting and fundamental work has be done by Krichever, Novikov, Schlichenmaier, and Sheinman on the representation theory of these algebras. In particular Wess–Zumino–Witten–Novikov theory and analogues of the Knizhnik–Zamolodchikov equations are developed for these algebras (see the survey article [24], and for example [26,23,21,22,25]). In [4] the authors gave commutations relations of the universal central extension of the DJKM Lie algebra in terms of a basis of the algebra and certain polynomials. More precisely in order to pin
*
Corresponding author. E-mail addresses: [email protected] (B. Cox), [email protected] (V. Futorny), [email protected] (J.A. Tirao).
0022-0396/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jde.2013.07.020
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
2847
down this central extension, we needed to describe four families of polynomials that appeared as coefficients in the commutator formulae. In this previous work we gave recursion relations for these polynomials and then found generating functions for them. Two of these families of polynomials are given in terms of elliptic integrals and the other two families are slight variations of ultraspherical polynomials. The main purpose of this note is to describe fourth order linear differential equations satisfied by the these two elliptic families of polynomials (see (3.22) and (3.24)), and to explain why these polynomials are orthogonal and non-classical (see Corollary 4.0.2, Theorem 5.0.5, and Theorem A.0.6). In fact one of the families are particular examples of associated ultraspherical polynomials (see [2]). The associated ultraspherical polynomials in turn are, up to factors of ascending factorials, particular associated Jacobi polynomials. The associated Jacobi polynomials are known to satisfy certain fourth order linear differential equations (see [13] and formula (48) in [30]). The proof given in the previous cited paper is obtained using MACSYMA and ideas coming from the classic text [29]. We show how we obtained a proof in the ultraspherical case that is done by hand and omit the calculation done for the remaining family of polynomials as it is similar. The referee was instrumental in identifying one of the families of non-classical polynomials describing the universal central extension and the references about associated ultraspherical and Jacobi polynomials. We would like to thank the referee for directing us to the relevant articles and books. For other examples of families of orthogonal polynomials that satisfy fourth order linear differential equations see the work of Samuel Shore, Allan Krall, and H.L. Krall [20,18,17]. The authors would like to thank Lance Littlejohn for these references and helpful correspondence. Still for even more examples of families of orthogonal polynomials that satisfy fourth order linear differential equations appearing in the study of supersingular j-invariants see the work of Kaneko and Zagier [19]. We plan to use these families of polynomials to describe free field realizations of the DJKM algebra in the setting of conformal field theory. In the case of affine Kac–Moody algebras, initial motivation for the use of Wakimoto’s realization (or free field realization) was to prove a conjecture of V. Kac and D. Kazhdan on the character of certain irreducible representations of affine Kac–Moody algebras at the critical level (see [28] and [10]). Another motivation for constructing free field realizations is that they are used to provide integral solutions to the KZ-equations (see for example [27] and [7] and their references). A third is that they are used to help in determining the center of a certain completion of the enveloping algebra of an affine Lie algebra at the critical level which is an important ingredient in the geometric Langland’s correspondence [11]. Yet a fourth is that free field realizations of an affine Lie algebra appear naturally in the context of the generalized AKNS hierarchies [9]. 2. DJKM algebras Let R be a commutative algebra defined over C. Consider the left R-module with action f ( g ⊗ h) = f g ⊗ h for f , g , h ∈ R and let K be the submodule generated by the elements 1 ⊗ f g − f ⊗ g − g ⊗ f . Then Ω R1 = F / K is the module of Kähler differentials. The element f ⊗ g + K is traditionally denoted by f dg. The canonical map d : R → Ω R1 by df = 1 ⊗ f + K . The exact differentials are the elements of the subspace dR. The coset of f dg modulo dR is denoted by f dg. As C. Kassel showed the universal central extension of the current algebra g ⊗ R where g is a simple finite dimensional Lie algebra defined over C, is the vector space g = (g ⊗ R ) ⊕ Ω R1 /dR with Lie bracket given by
[x ⊗ f , Y ⊗ g ] = [xy ] ⊗ f g + (x, y ) f dg ,
[ x ⊗ f , ω ] = 0,
ω , ω = 0,
where x, y ∈ g, and ω, ω ∈ Ω R1 /dR and (x, y ) denotes the Killing form on g. Consider the polynomial
p (t ) = t n + an−1 t n−1 + · · · + a0 , where ai ∈ C and an = 1. Fundamental to the description of the universal central extension for R = C[t , t −1 , u | u 2 = p (t )] are the following two results:
2848
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
Theorem 2.0.1. (See [3], Theorem 3.4.) Let R be as above. The set
t −1 dt , t −1 u dt , . . . , t −n u dt
forms a basis of Ω R1 /dR (omitting t −n u dt if a0 = 0). Proposition 2.0.2. (See [4], Lemma 2.0.2.) If um = p (t ) and R = C[t , t −1 , u | um = p (t )], then in Ω R1 /dR, one has n −1 (m + 1)n + im t n+i −1 u dt ≡ − (m + 1) j + mi a j t i + j −1 u dt mod dR .
(2.1)
j =0
In the Date–Jimbo–Miwa–Kashiwara setting one takes m = 2 and p (t ) = (t 2 − a2 )(t 2 − b2 ) = t 4 − (a2 + b2 )t 2 + (ab)2 with a = ±b and neither a nor b is zero. We fix from here onward R = C[t , t −1 , u | u 2 = (t 2 − a2 )(t 2 − b2 )]. As in this case a0 = (ab)2 , a1 = 0, a2 = −(a2 + b2 ), a3 = 0 and a4 = 1, then letting k = i + 3 the recursion relation in (2.1) looks like
(6 + 2k)t k u dt = −2(k − 3)(ab)2 t k−4 u dt + 2k a2 + b2 t k−2 u dt . √
After a change of variables, u → u /ab, t → t / ab, we may assume that a2 b2 = 1. Then the recursion relation looks like
(6 + 2k)t k u dt = −2(k − 3)t k−4 u dt + 4kct k−2 u dt ,
(2.2)
after setting c = (a2 + b2 )/2, so that p (t ) = t 4 − 2ct 2 + 1. Let P k := P k (c ) be the polynomial in c satisfy the recursion relation
(6 + 2k) P k (c ) = 4kc P k−2 (c ) − 2(k − 3) P k−4 (c ) for k 0. Then set
P (c , z) :=
P k (c ) zk+4 =
k−4
P k−4 (c ) zk ,
k0
so that after some straightforward rearrangement of terms we have
0=
(6 + 2k) P k (c ) zk − 4c
k0
k P k−2 (c ) zk + 2
k0
(k − 3) P k−4 (c ) zk
k0
d = −2z−4 + 8cz−2 − 6 P (c , z) + 2z−3 − 4cz−1 + 2z P (c , z)
+ 2z
−4
− 8cz
−2
dz
P −4 (c ) − 4c P −3 (c ) z
−1
− 2P −2 (c ) z−2 − 4P −1 (c ) z−1 .
Hence P (c , z) must satisfy the differential equation
d dz
P (c , z) −
3z4 − 4cz2 + 1 z5
− 2cz3
+z
P (c , z) =
2( P −1 + c P −3 ) z3 + P −2 z2 + (4cz2 − 1) P −4 z5 − 2cz3 + z
.
(2.3)
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
2849
This has integrating factor
μ(z) = exp
1 −2( z3 − cz) − dz z 1 − 2cz2 + z4
1 = exp − ln 1 − 2cz2 + z4 − ln( z) = √
1
z 1 − 2cz2 + z4
2
.
2.1. Elliptic case 1 If we take initial conditions P −3 (c ) = P −2 (c ) = P −1 (c ) = 0 and P −4 (c ) = 1 then we arrive at a generating function
P −4 (c , z) :=
P −4,k (c ) zk+4 =
k−4
P −4,k−4 (c ) zk
k0
defined in terms of an elliptic integral
P −4 (c , z) = z 1 − 2cz2 + z4
4cz2 − 1 z2 ( z4 − 2cz2 + 1)3/2
dz.
2.2. Elliptic case 2 If we take initial conditions P −4 (c ) = P −3 (c ) = P −1 (c ) = 0 and P −2 (c ) = 1, we arrive at a generating function defined in terms of another elliptic integral:
P −2 (c , z) = z 1 − 2cz2 + z4
1
( z4 − 2cz2 + 1)3/2
dz.
2.3. Gegenbauer case 3 If we take P −1 (c ) = 1, and P −2 (c ) = P −3 (c ) = P −4 (c ) = 0 and set P −1 (c , z) = then
P −1 (c , z) = (−1/2)
where Q n
1
3
c2 − 1
2 3
cz − z − cz + c z −
∞
n0
P −1,n−4 zn ,
(−1/2)
cQn
(c ) z
2n+1
,
k =2
(c ) is the n-th Gegenbauer polynomial. Hence P −1,−4 (c ) = P −1,−3 (c ) = P −1,−2 (c ) = P −1,2m (c ) = 0, P −1,−1 (c ) = 1, P −1,2n−3 (c ) = (−1/2)
for m 0 and n 2. The Q n
1 − c2
−c Q n (c ) , c2 − 1
(c ) are known to satisfy the second order differential equation:
d2 d2 c
(−1/2)
Qn
(−1/2)
(c ) + n(n − 1) Q n
(c ) = 0,
so that the P −1,k := P −1,k (c ) satisfy the second order differential equation
2850
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
c4 − c2
d2 d2 c
P −1,2n−3 + 2c c 2 + 1
d dc
P −1,2n−3 + −c 2 n(n − 1) − 2 P −1,2n−3 = 0
for n 2. 2.4. Gegenbauer case 4 Next we consider the initial conditions P −1 (c ) = 0 = P −2 (c ) = P −4 (c ) = 0 with P −3 (c ) = 1 and set
P −3 (c , z) =
P −3,n−4 (c ) z =
n0
(−1/2)
where Q n
1
n
2
c2 − 1
3
3
c z − cz − z + cz −
∞
(−1/2)
Qn
(c ) z
2n+1
,
k =2
(c ) is the n-th Gegenbauer polynomial. Hence P −3,−4 (c ) = P −3,−2 (c ) = P −3,−1 (c ) = P −1,2m (c ) = 0, P −3,−3 (c ) = 1, P −3,2n−3 (c ) =
− Q n (c ) , c2 − 1
for m 0 and n 2 and hence
c2 − 1
d2 d2 c
P −3,2n−3 + 4c
d dc
P −3,2n−3 − (n + 1)(n − 2) P −3,2n−3 = 0
for n 2 and P −1,2n−3 = c P −3,2n−3 for n 2. We’ll see in the last section how the polynomials described in the elliptic case 1 are particular examples of associated ultraspherical polynomials. The importance of these families of polynomials come from our previous work describing the universal central extension of the DJKM algebra: Theorem 2.4.1. (See [4].) Let g be a simple finite dimensional Lie algebra over the complex numbers with the Killing form ( | ) and define ψi j (c ) ∈ Ω R1 /dR by
ψi j (c ) =
⎧ω i + j −2 ⎪ ⎪ ⎪ ⎨ P −3,i + j −2 (c )(ω−3 + c ω−1 )
for i + j = 1, 0, −1, −2,
for i + j = 2n − 1 3, n ∈ Z, ⎪ P ( c )( c ω + ω ) for i + j = −2n + 1 −3, n ∈ Z, −3 −1 −3 , i + j −2 ⎪ ⎪ ⎩ P −4,|i + j |−2 (c )ω−4 + P −2,|i + j |−2 (c )ω−2 for |i + j | = 2n 2, n ∈ Z.
(2.4)
The universal central extension of the Date–Jimbo–Kashiwara–Miwa algebra is the Z2 -graded Lie algebra
g = g0 ⊕ g1 , where
g0 = g ⊗ C t , t −1 ⊕ Cω0 , with bracket
g1 = g ⊗ C t , t −1 u ⊕ Cω−4 ⊕ Cω−3 ⊕ Cω−2 ⊕ Cω−1
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
x ⊗ t i , y ⊗ t j = [x, y ] ⊗ t i + j + δi + j ,0 j (x, y )ω0 ,
x ⊗ t i −1 u , y ⊗ t j −1 u = [x, y ] ⊗ t i + j +2 − 2ct i + j + t i + j −2
2851
+ δi + j ,−2 ( j + 1) − 2c j δi + j ,0 + ( j − 1)δi + j ,2 (x, y )ω0 ,
x ⊗ t i −1 u , y ⊗ t j = [x, y ]u ⊗ t i + j −1 + j (x, y )ψi j (c ).
3. Differential equations for elliptic type 1 and 2 3.1. Elliptic type 1 How we arrive at the fourth order linear differential equation that the polynomials P −4,n satisfy stems from the approach used in Afken’s book [1] for finding the second order linear differential equation that Legendre polynomials satisfy using only the recursion relation they satisfy. His technique seems to be indirectly based on ideas used in the theory of Gröbner basis. Unfortunately the calculations and relations involved are rather tedious. From now on we are going to reindex the polynomials P −4,n :
P −4 (c , z) = z 1 − 2cz2 + z4
4cz2 − 1
dz = z2 ( z4 − 2cz2 + 1)3/2 1
∞
P −4,n (c ) zn
n =0
16 2 = 1 + z4 + z6 + 32c 2 − 5 z8 + c 8c − 3 z10 5 35 105 4c
−
(2048c 4 − 1248c 2 + 75) 1155
z12 + O z14 .
This means that now P −4,0 (c ) = 1, P −4,1 (c ) = P −4,2 (c ) = P −4.3 (c ) = 0. Besides P −4,0 (c ), the first few nonzero polynomials in c are
P −4,4 (c ) = 1, P −4,10 =
P −4,6 =
16 2 c 8c − 3 , 105
4c 5
,
P −4,12 = −
P −4,8 =
32c 2 − 5 35
,
(2048c 4 − 1248c 2 + 75) 1155
and P −4,n (c ) satisfy the following recursion:
(6 + 2k) P k+4 (c ) = 4kc P k+2 (c ) − 2(k − 3) P k (c ).
(3.1)
Our goal in this section is to find families of linear differential equation in c that these polynomials satisfy. We start off with the generating function
P −4 (c , z) = z 1 − 2cz2 + z4
=z
1 − 2cz2
+
z4
4cz2 − 1 z2 ( z4 − 2cz2 + 1)3/2 ∞ (3/2) 4c Q n (c ) n =0
(λ)
2n + 1
z
2n+1
dz
∞ (3/2) Qn (c ) 2n−1 − z , 2n − 1 n =0
where Q n (c ) is the n-Gegenbauer polynomial. These polynomials satisfy the second order linear ODE:
2852
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
1 − c 2 y − (2λ + 1)c y + n(n + 2λ) y = 0,
where the derivative is with respect to c. Thus for λ = 3/2 we get
1 − c2
(3/2)
Qn
(3/2) (3/2) (c ) − 4c Q n (c ) + n(n + 3) Q n (c ) = 0.
(3.2)
Rewrite the expansion formula for P −4 (c , z) to get
z−1 1 − 2cz2 + z4
−1/2
P −4 (c , z) =
∞ (3/2) 4c Q n (c )
2n + 1
n =0
z2n+1 −
∞ (3/2) Qn (c ) 2n−1 , z 2n − 1
(3.3)
n =0
2
d d and apply the differential operator L := (1 − c 2 ) dc 2 − 4c dc to the right hand side to get
(3/2)
L 4c Q n
d2 d (3/2) (c ) = 1 − c 2 − 4c 4c Q n (c ) 2 dc
dc
(3/2) = −8(n + 1) Q n+1 (c ) − 4c n2
(3/2) + n − 2 Q n (c )
using the identity
1 − c2
d dc
(λ)
(λ)
(λ)
Q n (c ) = (n + 2λ)c Q n (c ) − (n + 1) Q n+1 (c ).
(3.4)
Using a number of simplifications that we have put on http://arxiv.org/archive/math we get
L
∞ (3/2) 4c Q n (c ) n =0
2n + 1
z
2n+1
=
(3/2) (3/2) ∞ −8(n + 1) Q n+1 (c ) − 4c (n2 + n − 2) Q n (c )
2n + 1
n =0
= −4
1 z( z4 − 2cz2 + 1)3/2
− cz2
dz
+ 9c
d
1
−4
z2 ( z4 − 2cz2 + 1)3/2
1
( z4 − 2cz2 + 1)3/2 1
( z4 − 2cz2 + 1)3/2
z2n+1
−
dz
cz
( z4 − 2cz2 + 1)3/2
dz.
In addition we have
∞ ∞ (3/2) (3/2) Qn (c ) 2n−1 n(n + 3) Q n (c ) 2n−1 L z z =− 2n − 1 2n − 1 n =0
n =0
∞ (3/2) 1 d2 9 d 7 Qn (c ) 2n−1 = − z2 2 + z + z 4 dz 4 dz 4 2n − 1 n =0 1 d 1 9 − = − z2 4 dz z2 ( z4 − 2cz2 + 1)3/2 4z( z4 − 2cz2 + 1)3/2
7 1 − dz. 4 z2 ( z4 − 2cz2 + 1)3/2
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
2853
Thus the right hand side of (3.3) becomes
−4
1 z( z4 − 2cz2 + 1)3/2
d
− cz2
dz
=− − +
1
−
( z4 − 2cz2 + 1)3/2
( z4 − 2cz2 + 1)3/2
d
1
4
dz
z2 ( z4 − 2cz2 + 1)3/2
7
4
1 z2 ( z4 − 2cz2 + 1)3/2 7
4z( z4 − 2cz2 + 1)3/2 cz
( z4 − 2cz2 + 1)3/2 9
1
√
4 z z4 − 2cz2 + 1
+
−
cz
( z4 − 2cz2 + 1)3/2
dz
1
+ z2
dz
z2 ( z4 − 2cz2 + 1)3/2
1
+ 9c
+
1
−4
+
9 4z( z4 − 2cz2 + 1)3/2
dz 6cz2 ( z2 − c )
( z4 − 2cz2 + 1)5/2 2z2 (4z4 − 5cz2 + 1)
z3 (−2cz2 + z4 + 1)5/2
P −4 (c , z).
Applying the differential operator L to the left hand side of (3.3), we get
L z−1 1 − 2cz2 + z4
−1/2
P −4 (c , z)
d 2 −1 −1/2 −1/2 d −1 z 1 − 2cz2 + z4 P −4 (c , z) − 4c z 1 − 2cz2 + z4 P −4 (c , z) = 1 − c2 2 dc
dc
=
z(−4c + (3 + 5c ) z − 4cz ) 2
2
(1 − 2cz2 + z4 )5/2 +
4
P −4 (c , z) +
−4c + (2 + 6c 2 ) z2 − 4cz4 d P −4 (c , z) dc z(1 − 2cz2 + z4 )3/2
(1 − c 2 ) d2 P −4 (c , z) . z(1 − 2cz2 + z4 )1/2 dc 2
Hence we have
z(−4c + (3 + 5c 2 ) z2 − 4cz4 )
(1 − 2cz2
+
z4 )5/2
P −4 (c , z) +
−4c + (2 + 6c 2 ) z2 − 4cz4 d P −4 (c , z) 2 4 3 / 2 dc z(1 − 2cz + z )
(1 − c 2 ) d2 P −4 (c , z) z(1 − 2cz2 + z4 )1/2 dc 2 −4cz2 − 7 1 2 d = − cz dz ( z4 − 2cz2 + 1)3/2 4z( z4 − 2cz2 + 1)3/2 1 1 d 1 9 + c √ + z2 P −4 (c , z) 2 4 2 3 / 2 4 4 dz z ( z − 2cz + 1) 4 z z − 2cz2 + 1 +
as
−3/2 −2(4z4 − 5cz2 + 1) d −2 4 = 3 4 z z − 2cz2 + 1 . dz z ( z − 2cz2 + 1)5/2
2854
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
As a consequence
z(−4c + (3 + 5c 2 ) z2 − 4cz4 )
(1 − 2cz2
+
z4 )5/2
P −4 (c , z) +
−4c + (2 + 6c 2 ) z2 − 4cz4 d P −4 (c , z) 2 4 3 / 2 dc z(1 − 2cz + z )
d2 (1 − c 2 ) P −4 (c , z) z(1 − 2cz2 + z4 )1/2 dc 2 4cz2 + 7 6z3 (c − z2 ) − c =− 4z( z4 − 2cz2 + 1)3/2 ( z4 − 2cz2 + 1)5/2 +
−
2(4z4 − 5cz2 + 1) z( z4
− 2cz2
+ 1)5/2
+
9
1
√
4 z z4 − 2cz2 + 1
P −4 (c , z),
which gives us
9
− + 5cz2 − 4
= −
9 4
15 4
+ 4c 2 z4 + 5cz6
z4 − 2cz2 + 1
2
+ z2 −4c + 3 + 5c 2 z2 − 4cz4 P −4 (c , z)
d P −4 (c , z) + −4c + 2 + 6c 2 z2 − 4cz4 z4 − 2cz2 + 1 dc
2 d P −4 (c , z) . + 1 − c 2 z4 − 2cz2 + 1 2 2
dc
Expanding this out in detail and writing P −4,k (c ) as P k we will obtain
0 = −9P n + 20c P n−2 − 16c 2 + 6 P n−4 + 20c P n−6 − 9P n−8
− 16c P n + 8 1 + 7c 2 P n −2 − 48c 1 + c 2 P n −4 + 8 1 + 7c 2 P n −6 − 16c P n −8 + 4 1 − c 2 P n − 4c P n−2 + 2 1 + 2c 2 P n−4 − 4c P n−6 + P n−8 .
(3.5)
We now differentiate with respect to c the recursion relations
(6 + 2k) P k+4 = 4kc P k+2 − 2(k − 3) P k
(3.6)
to get
(6 + 2k) P k +4 = 4k P k+2 + 4kc P k +2 − 2(k − 3) P k ,
(3.7)
(6 + 2k) P k+4
(3.8)
= 8k P k+2 + 4kc P k+2
− 2(k − 3) P k .
After setting k = n − 8 in the last equation we get
0 = −(2n − 10) P n−4 + 8(n − 8) P n −6 + 4(n − 8)c P n−6 − 2(n − 11) P n−8 . So multiplying (3.5) by (n − 11) and adding it to 2(1 − c 2 ) times the above gives us
(3.9)
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
2855
0 = (n − 11) −9P n + 20c P n−2 − 16c 2 + 6 P n−4 + 20c P n−6 − 9P n−8
− 16c P n + 8 1 + 7c 2 P n −2 − 48c 1 + c 2 P n −4 − 16c P n −8 + 4 1 − c 2 P n − 4c P n−2 + 8 c 2 (5n − 61) + (3n − 27) P n −6 − 4 1 − c 2 4c 2 (n − 11) + n − 17 P n−4 + 8(n − 14)c 1 − c 2 P n−6 .
(3.10)
Setting k = n − 8 in (3.7) we get
0 = −(2n − 10) P n −4 + 4(n − 8) P n−6 + 4(n − 8)c P n −6 − 2(n − 11) P n −8 .
(3.11)
Multiplying this equation by −8c and add to the previous equation we get
0 = (n − 11) −9P n + 20c P n−2 − 16c 2 + 6 P n−4 − 9P n−8
− 16c P n + 8 1 + 7c 2 P n −2 + 4 1 − c 2 P n − 4c P n−2 + 16c −3 c 2 + 1 (n − 11) + n − 5 P n −4 − 4 1 − c 2 4c 2 (n − 11) + n − 17 P n−4 − 12c (n − 3) P n−6 + 8 c 2 (n − 29) + 3(n − 9) P n −6 + 8(n − 14)c 1 − c 2 P n−6 .
Finally if we multiply the previous equation by 2 and add it to −9 times (3.6) we get
0 = 2(n − 11) −9P n + 20c P n−2 − 16c P n + 8 1 + 7c 2 P n −2 + 4 1 − c 2
P n − 4c P n−2
+ 6(n + 7) − 32c 2 (n − 11) P n−4 + 32c −3 c 2 + 1 (n − 11) + n − 5 P n −4 − 8 1 − c 2 4c 2 (n − 11) + n − 17 P n−4 − 60c (n − 6) P n−6 + 16 c 2 (n − 29) + 3(n − 9) P n −6 + 16(n − 14)c 1 − c 2 P n−6 .
This is an equation without the P n−8 in it. We now get rid of the P n−6 ’s in them. After setting k = n − 6 in (3.8) we get
0 = −(2n − 6) P n−2 + 8(n − 6) P n −4 + 4(n − 6)c P n−4 − 2(n − 9) P n−6 .
(3.12)
Now we multiply the previous equation by −8(n − 14)c (1 − c 2 ) and add it to n − 9 times the equation before it, we obtain
0 = 2(n − 11)(n − 9) −9P n + 20c P n−2 − 16c P n + 8 1 + 7c 2 P n −2 + 4 1 − c 2 P n
+ (n − 9) 6(n + 7) − 32c 2 (n − 11) P n−4 + 16c c 2 − 1 n2 − 23n + 156 P n−2 − 32c c 2 n2 − 20n + 129 + 4n2 − 86n + 420 P n −4 − 8 c 2 − 1 60c 2 + n2 − 26n + 153 P n−4 − 60c (n − 9)(n − 6) P n−6 + 16(n − 9) c 2 (n − 29) + 3(n − 9) P n −6 .
From (3.7) with k = n − 6 one has
0 = −(2n − 6) P n −2 + 4(n − 6) P n−4 + 4(n − 6)c P n −4 − 2(n − 9) P n −6 .
(3.13)
2856
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
We multiply this equation by 8(c 2 (n − 29) + 3(n − 9)) and add it to the previous equation to get
0 = 2(n − 11)(n − 9) −9P n + 20c P n−2 − 16c P n + 8 1 + 7c 2 P n −2 + 4 1 − c 2 P n
− 8(2n − 6) c 2 (n − 29) + 3(n − 9) P n −2 + 16c c 2 − 1 n2 − 23n + 156 P n−2 − 6 80c 2 (n − 5) − 17n2 + 242n − 801 P n−4 − 32c 15c 2 (n − 3) + n2 − 41n + 258 P n −4 − 8 c 2 − 1 60c 2 + n2 − 26n + 153 P n−4 − 60c (n − 9)(n − 6) P n−6 .
From (3.6) with k = n − 6 one has
0 = −(2n − 6) P n−2 + 4(n − 6)c P n−4 − 2(n − 9) P n−6 . We multiply this equation by −30c (n − 6) and add it to the previous equation to get
0 = 2(n − 11)(n − 9) −9P n − 16c P n + 8 1 + 7c 2 P n −2 + 4 1 − c 2 P n
+ 20c 5n2 − 67n + 252 P n−2 − 8(2n − 6) c 2 (n − 29) + 3(n − 9) P n −2 + 16c c 2 − 1 n2 − 23n + 156 P n−2 + 6 −20c 2 (n − 4)2 + 17n2 − 242n + 801 P n−4 − 32c 15c 2 (n − 3) + n2 − 41n + 258 P n −4 − 8 c 2 − 1 60c 2 + n2 − 26n + 153 P n−4 .
The above equation now does not have the index n − 6 in it. Now we want to eliminate the indices with n − 4 in them. From (3.8) with k = n − 4 one has
0 = −(2n − 2) P n + 8(n − 4) P n −2 + 4(n − 4)c P n−2 − 2(n − 7) P n−4 . We multiply this equation by −4(c 2 − 1)(60c 2 + n2 − 26n + 153) and add it to n − 7 times the previous equation to get
0 = 2(n − 11)(n − 9)(n − 7) −9P n − 16c P n
+ 20c (n − 7) 5n2 − 67n + 252 P n−2 + 6(n − 7) −20c 2 (n − 4)2 + 17n2 − 242n + 801 P n−4 − 32c (n − 7) 15c 2 (n − 3) + n2 − 41n + 258 P n −4 + 480 c 2 − 1 c 2 (n − 1) − n + 9 P n − 32 60c 4 (n − 4) + c 2 −2n3 + 45n2 − 484n + 1749 + 15 n2 − 14n + 45 P n −2 − 960c c 2 − 1 c 2 (n − 4) − n + 8 P n−2 .
From (3.7) with k = n − 4 one has
0 = −(2n − 2) P n + 4(n − 4) P n−2 + 4(n − 4)c P n −2 − 2(n − 7) P n −4 . We multiply this equation by −16c (15c 2 (n − 3) + n2 − 41n + 258) and add it to the previous equation to get
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
2857
0 = −18(n − 11)(n − 9)(n − 7) P n
+ 480c c 2 n2 − 4n + 3 − n2 + 4n + 29 P n + 480 c 2 − 1 c 2 (n − 1) − n + 9 P n + 12c −80c 2 n2 − 7n + 12 + 3n3 + 70n2 − 1049n + 2564 P n−2 − 480 2c 4 n2 − 5n + 4 − 3c 2 n2 − 8n + 7 + n2 − 14n + 45 P n −2 − 960c c 2 − 1 c 2 (n − 4) − n + 8 P n−2 + 6(n − 7) −20c 2 (n − 4)2 + 17n2 − 242n + 801 P n−4 .
From (3.6) with k = n − 4 one has
0 = −2(n − 1) P n + 4(n − 4)c P n−2 − 2(n − 7) P n−4 . We multiply this equation by 3(−20c 2 (n − 4)2 + 17n2 − 242n + 801) and add it to the previous equation to get
0 = 120(n − 4)2 c 2 (n − 1) − n + 9 P n
+ 480c c 2 n2 − 4n + 3 − n2 + 4n + 29 P n + 480 c 2 − 1 c 2 (n − 1) − n + 9 P n − 240c (n − 2)2 c 2 (n − 4) − n + 8 P n−2 − 480 2c 4 n2 − 5n + 4 − 3c 2 n2 − 8n + 7 + n2 − 14n + 45 P n −2 − 960c c 2 − 1 c 2 (n − 4) − n + 8 P n−2 .
This can be rewritten as
0 = (n − 4)2 c 2 (n − 1) − n + 9 P n
+ 4c c 2 n2 − 4n + 3 − n2 + 4n + 29 P n + 4 c 2 − 1 c 2 (n − 1) − n + 9 P n − 2c (n − 2)2 c 2 (n − 4) − n + 8 P n−2 − 4 2c 4 n2 − 5n + 4 − 3c 2 n2 − 8n + 7 + n2 − 14n + 45 P n −2 − 8c c 2 − 1 c 2 (n − 4) − n + 8 P n−2 .
(3.14)
We have now reduced to a differential equation with only the indices n and n − 2. There is a bit of a trick to reduce it down to a linear ODE with only the index n in it. The somewhat vague idea is to find more equations to in order to cancel all but terms with the index n in them. From (3.8) with k = n − 2 one has
0 = −2(n + 1) P n+2 + 8(n − 2) P n + 4(n − 2)c P n − 2(n − 5) P n−2 . We multiply this equation by −4c (c 2 − 1)(c 2 (n − 4) + 8 − n) and add it to n − 5 times the previous equation to get
2858
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
0 = −2(n − 5)c (n − 2)2 c 2 (n − 4) − n + 8 P n−2
− 4(n − 5) 2c 4 n2 − 5n + 4 − 3c 2 n2 − 8n + 7 + n2 − 14n + 45 P n −2 + (n − 5)(n − 4)2 c 2 (n − 1) − n + 9 P n + 4c −8c 4 n2 − 6n + 8 + c 2 n3 + 7n2 − 105n + 177 − n3 + n2 + 89n − 273 P n − 4 c 2 − 1 4c 4 n2 − 6n + 8 + c 2 −5n2 + 46n − 69 + n2 − 14n + 45 P n + 8(n + 1)c c 2 − 1 c 2 (n − 4) + 8 − n (n + 1) P n+2 .
From (3.7) with k = n − 2 one has
0 = −2(n + 1) P n +2 + 4(n − 2) P n + 4(n − 2)c P n − 2(n − 5) P n −2 . We multiply this equation by
−2 2c 4 (n − 4)(n − 1) − 3c 2 (n − 7)(n − 1) + (n − 9)(n − 5) and add it to the previous equation to get
0 = −2(n − 5)c (n − 2)2 c 2 (n − 4) − n + 8 P n−2
+ −16c 4 n3 − 7n2 + 14n − 8 + c 2 n4 + 10n3 − 171n2 + 416n − 256 − n2 n2 − 14n + 45 P n − 4c (n + 1) 4c 4 n2 − 6n + 8 + c 2 −7n2 + 60n − 93 + 3 n2 − 12n + 31 P n − 4 c 2 − 1 4c 4 n2 − 6n + 8 + c 2 −5n2 + 46n − 69 + n2 − 14n + 45 P n + 4 2c 4 (n − 4)(n − 1) − 3c 2 (n − 7)(n − 1) + (n − 9)(n − 5) (n + 1) P n +2 + 8(n + 1)c c 2 − 1 c 2 (n − 4) + 8 − n (n + 1) P n+2 .
Next we get rid of the P n−2 term by multiplying (3.6) with k = n − 2;
0 = −2(n + 1) P n+2 + 4(n − 2)c P n − 2(n − 5) P n−2 , by −c (n − 2)2 (c 2 (n − 4) − n + 8) and adding it to the previous equation:
0 = −n2 4c 4 n2 − 6n + 8 + c 2 −5n2 + 46n − 69 + n2 − 14n + 45 P n
− 4c (n + 1) 4c 4 n2 − 6n + 8 + c 2 −7n2 + 60n − 93 + 3 n2 − 12n + 31 P n − 4 c 2 − 1 4c 4 n2 − 6n + 8 + c 2 −5n2 + 46n − 69 + n2 − 14n + 45 P n + 2c (n + 1)(n − 2)2 c 2 (n − 4) − n + 8 P n+2 + 4(n + 1) 2c 4 (n − 4)(n − 1) − 3c 2 (n − 7)(n − 1) + (n − 9)(n − 5) P n +2 + 8(n + 1)2 c c 2 − 1 c 2 (n − 4) + 8 − n P n+2 .
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
2859
Letting n → n − 2 in the above equation we get
0 = −(n − 2)2 4c 4 n2 − 10n + 24 + c 2 −5n2 + 66n − 181 + n2 − 18n + 77 P n−2
− 4c (n − 1) 4c 4 n2 − 10n + 24 + c 2 −7n2 + 88n − 241 + 3 n2 − 16n + 59 P n −2 − 4 c 2 − 1 4c 4 n2 − 10n + 24 + c 2 −5n2 + 66n − 181 + n2 − 18n + 77 P n−2 + 2c (n − 1)(n − 4)2 c 2 (n − 6) − n + 10 P n + 4(n − 1) 2c 4 (n − 6)(n − 3) − 3c 2 (n − 9)(n − 3) + (n − 11)(n − 7) P n + 8(n − 1)2 c c 2 − 1 c 2 (n − 6) − n + 10 P n . (3.15)
We compare this to (3.14)
0 = −2c (n − 2)2 c 2 (n − 4) − n + 8 P n−2
− 4 2c 4 n2 − 5n + 4 − 3c 2 n2 − 8n + 7 + n2 − 14n + 45 P n −2 − 8c c 2 − 1 c 2 (n − 4) − n + 8 P n−2 + (n − 4)2 c 2 (n − 1) − n + 9 P n + 4c c 2 n2 − 4n + 3 − n2 + 4n + 29 P n + 4 c 2 − 1 c 2 (n − 1) − n + 9 P n .
(3.16)
Our goal is to eliminate the P n−2 term, then the P n −2 and finally the P n−2 to arrive at a differential equation with just the derivatives of P n in it. We first have to lower the degree of c in the polynomial in front of P n−2 . Thus if we multiply (3.16) by −2c (n − 6) and add it to (3.15) we get
0 = (n − 11)(n − 2)2 c 2 (n + 1) − n + 7 P n−2
+ 4c (n − 11) c 2 n2 − 1 − n2 + 33 P n −2 + 4(n − 11) c 2 − 1 c 2 (n + 1) − n + 7 P n−2 − 8c (n − 11)(n − 4)2 c 2 (n − 7) − n − 1 P n − 4(n − 11) c 2 n2 − 8n + 39 − n2 + 8n − 7 P n − 32c (n − 11) c 2 − 1 P n .
If n = 11 we get
0 = (n − 2)2 c 2 (n + 1) − n + 7 P n−2
+ 4c c 2 n2 − 1 − n2 + 33 P n −2 + 4 c 2 − 1 c 2 (n + 1) − n + 7 P n−2 − 8c (n − 4)2 P n − 4 c 2 n2 − 8n + 39 − n2 + 8n − 7 P n − 32c c 2 − 1 P n .
(3.17)
2860
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
We lower the degree of c in the polynomial in front of P n−2 another time by multiplying this last equation by 2c (n − 4) and add it to n + 1 times (3.14) to get
0 = 8c (n − 9)(n − 2)2 P n−2
+ 4(n − 9) c 2 n2 − 4n + 27 − n2 + 4n + 5 P n −2 + 32(n − 9)c c 2 − 1 P n−2 + (n − 9)(n − 4)2 c 2 (n − 7) − n − 1 P n − 4c (n − 9) c 2 n2 − 12n + 35 − n2 + 12n − 3 P n + 4(n − 9) c 2 − 1 c 2 (n − 7) − n − 1 P n .
So that if n = 9, 11 one has
0 = 8c (n − 2)2 P n−2
+ 4 c 2 n2 − 4n + 27 − n2 + 4n + 5 P n −2 + 32c c 2 − 1 P n−2 + (n − 4)2 c 2 (n − 7) − n − 1 P n − 4c c 2 n2 − 12n + 35 − n2 + 12n − 3 P n + 4 c 2 − 1 c 2 (n − 7) − n − 1 P n .
(3.18)
We lower the degree of c in the polynomial in front of P n−2 another time by multiplying this last equation by c (n + 1) and add it to −8 times (3.17) to get
0 = 8(n − 7)(n − 2)2 P n−2
+ 4c (n − 7) c 2 n2 − 4n − 5 − n2 + 4n + 37 P n −2 + 32 c 2 − 1 (n − 7) P n−2 + c (n − 7)(n − 4)2 c 2 (n + 1) − n − 9 P n − 4(n − 7) c 4 n2 − 4n − 5 − c 2 n2 + 4n − 45 + 8(n − 1) P n + 4c c 2 − 1 (n − 7) c 2 (n + 1) − n − 9 P n .
Which if n = 7, 9, 11, then we have
0 = 8(n − 2)2 P n−2
+ 4c c 2 n2 − 4n − 5 − n2 + 4n + 37 P n −2 + 32 c 2 − 1 P n−2 + c (n − 4)2 c 2 (n + 1) − n − 9 P n − 4 c 4 n2 − 4n − 5 − c 2 n2 + 4n − 45 + 8(n − 1) P n + 4c c 2 − 1 c 2 (n + 1) − n − 9 P n .
(3.19)
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
2861
We now want to get rid of the term with P n−2 in it. This is done by multiplying the previous equation by c and add it to −1 times (3.18) we get
2
0 = 4 c 2 − 1 (n − 5)(n + 1) P n −2
2 + c 2 − 1 (n − 4)2 (n + 1) P n 2 − 4c c 2 − 1 (n − 5)(n + 1) P n 3 + 4 c 2 − 1 (n + 1) P n .
Thus as c = ±1 and we are assuming n = −1, 7, 9, 11 then we have
0 = 4(n − 5) P n −2 + (n − 4)2 P n − 4c (n − 5) P n + 4 c 2 − 1 P n .
(3.20)
If we differentiate this with respect to c we get
0 = 4(n − 5) P n−2 + (n − 4)2 P n − 4(n − 5) P n − 4c (n − 5) P n + 8c P n + 4 c 2 − 1 P n
= 4(n − 5) P n−2 + (n − 6)2 P n − 4c (n − 7) P n + 4 c 2 − 1 P n .
(3.21)
Now we work on the coefficient in front of P n −2 to eliminate it. We can multiply the previous equation by −8(c 2 − 1) and add to n − 5 times (3.19) to give us
0 = 8(n − 5)(n − 2)2 P n−2
+ 4c (n − 5) c 2 n2 − 4n − 5 − n2 + 4n + 37 P n −2 + c (n − 5)(n − 4)2 c 2 (n + 1) − n − 9 P n + 4 c 4 −(n − 5)2 (n + 1) + c 2 n3 − 3n2 − 41n + 153 − 6n2 + 24n + 32 P n + 4c c 2 − 1 8(n − 7) + (n − 5) c 2 (1 + n) − 9 − n P n 2 − 32 c 2 − 1 P n .
Now we multiply (3.20) by c (37 + 4n − n2 + c 2 (n − 5)(n + 1)) and add it to −1 times the equation above to give us
0 = −8(n − 5)(n − 2)2 P n−2 + 8c (n − 4)2 (n − 1) P n
2 − 8 c 2 − 1 3n2 − 12n − 16 P n + 192c c 2 − 1 P n + 32 c 2 − 1 P n .
Thus if c = ±1, n = −1, 7, 9, 11 we have
0 = −(n − 5)(n − 2)2 P n−2 + c (n − 4)2 (n − 1) P n
2 − c 2 − 1 3n2 − 12n − 16 P n + 24c c 2 − 1 P n + 4 c 2 − 1 P n .
2862
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
If we differentiate this with respect to c we get
0 = −(n − 5)(n − 2)2 P n −2
+ (n − 4)2 (n − 1) P n + c n3 − 15n2 + 48n + 16 P n + −c 2 3n2 − 12n − 88 + 3n2 − 12n − 40 P n + 40c c 2 − 1 P n 2 ( i v ) + 4 c2 − 1 P n . Now we multiply this by 4 and add it to (n − 2)2 times (3.20) to get the 4th order linear differential equation satisfied by the polynomials P n . We proved Theorem 3.1.1. The polynomials P n = P −4,n satisfy the following differential equation:
+ 160c c 2 − 1 P n − 8 c 2 n2 − 4n − 46 − n2 + 4n + 22 P n − 24c n2 − 4n − 6 P n + (n − 4)2n2 P n = 0.
16 c 2 − 1
2
(i v )
Pn
(3.22)
3.2. Elliptic case 2 From now on we are going to reindex the polynomials P −2,n :
P −2 (c , z) = z 1 − 2cz2 + z4
1
( z4 − 2cz2 + 1)3/2
=
∞
P −2,n (c ) zn .
n =0
This means that now P −2,2 (c ) = 1, P −2,3 (c ) = P −2,1 (c ) = P −2,0 (c ) = 0 and P −2,n (c ) satisfy the following recursion:
(6 + 2k) P k+4 (c ) = 4kc P k+2 (c ) − 2(k − 3) P k (c ).
(3.23)
The first few nonzero polynomials in c are
P −2,2 (c ) = 1,
P −2,6 = 1/5,
P −2,10 = −7 + 32c 2 /105,
P −2,8 = 8c /35,
P −2,12 = 8c −29 + 64c 2 /1155,
so that
P −2 (c , z) = z2 +
1 5
z6 +
8c 35
z8 +
32c 2 − 7 105
z10 +
8c (64c 2 − 29) 1155
z12 + O z14 .
After a very similar lengthy analysis as in the previous section we arrive at the following result: Theorem 3.2.1. The polynomials P n = P −2,n satisfy the following fourth order linear differential equation:
+ 160c c 2 − 1 P n − 8 c 2 n2 − 4n − 42 − n2 + 4n + 18 P n − 24c n2 − 4n − 2 P n + (n − 6)(n − 2)2 (n + 2) P n = 0.
16 c 2 − 1
2
(i v )
Pn
(3.24)
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
2863
4. Associated ultraspherical polynomials After shifting the indices back by 4 we obtain that both families of the polynomials P −4,n (c ) and P −2,n (c ) satisfy the recurrence relation
2kc P k−2 (c ) = (3 + k) P k (c ) + (k − 3) P k−4 (c ).
(4.1)
Note that all odd polynomials are zero. Set k = 2(n + 1) and q s := P 2s . Then we have
4(n + 1)cqn = (2n + 5)qn+1 + (2n − 1)qn−1 ,
(4.2)
where q s = q s (c ). For q s (c ) = P −4,2s (c ) one has
q−2 = P −4,−4 = 1, q1 = P −4,2 = 4c /5,
q−1 = P −4,−2 = 0,
q2 = P −4,4 =
32c − 5 2
35
,
q0 = P −4,0 = 1, q3 = P −4,6 =
16 2 c 8c − 3 . 105
(4.3)
We will show that they are special cases of the associated ultraspherical polynomials. (ν ) For c and ν two complex constants, if one lets C n (x; c ) denote the associated ultraspherical poly(ν ) (ν ) nomials with initial conditions C −1 (x; c ) = 0, C 0 (x; c ) = 1 then they are defined so as to satisfy the difference equation (ν )
(ν )
(ν )
2x(n + ν + c )C n (x; c ) = (n + c + 1)C n+1 (x; c ) + (2ν + n + c − 1)C n−1 (x; c ).
(4.4)
Then setting c = 3/2 and ν = −1/2 this equation is the same recurrence relation as (4.2). Hence, we can conclude that polynomials P −4,n is a special case of associated ultraspherical polynomials. (α ,β)
For α , β, c ∈ C constants, the associated Jacobi polynomials P n the recurrence relation
(x; c ) are defined so as to satisfy
2(n + c + 1)(n + c + γ )(2n + 2c + γ − 1) pn+1
= (2n + 2c + γ ) (2n + 2c + γ − 1)(2n + 2c + γ + 1)x + (γ − 1)(γ − 2β − 1) pn − 2(n + c + γ − β − 1)(n + c + β)(2n + 2c + γ + 1) pn−1 , (α ,β)
(4.5)
(α ,β)
where n ∈ N, γ = α + β + 1, P −1 (x, c ) = 0; P 0 (x, c ) = 1. One can check that the associated ultraspherical polynomials are related to the associated Jacobi polynomials through the formula (ν −1/2,ν −1/2)
Pn
(x; β) =
(ν + c + 12 )n (ν ) C n (x; β). (2ν + c )n
The weight measure for the associated ultraspherical polynomials is known (see [2]) and is given by
ρ (cos t ) =
(2 sin t )2β−1 [Γ (λ + γ )]2 2ı t −2 2 F 1 1 − β, γ ; γ + β; e 2π Γ (2β + γ )Γ (γ + 1)
(4.6)
2864
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
for 0 t π and where 2 F 1 (a, b; c ; z) is the hypergeometric function 2 F 1 (a, b ; c ; z ) =
∞ (a)n (b)n n =0
(c )n
zn .
Corollary 4.0.2. Polynomials P −4,n are non-classical orthogonal polynomials. Finally we should point out that Wimp (see [13], Eqs. 5.7.10–5.7.12 and formula (48) in [30]) using MACSYMA and ideas from [29], deduced that the associated Jacobi Polynomials satisfy the fourth order differential equation
A 0 (x) y ( vi ) + A 1 (x) y + A 2 (x) y + A 3 (x) y + A 4 (x) y = 0,
(4.7)
where
A 0 (x) = 1 − x2
2
,
A 1 (x) = 10x x2 − 1 ,
A 2 (x) = −(1 − x)2 2K + 2C + γ 2 − 25 + 2(1 − x)(2K + 2C + 2αγ ) + 2(α + 1) − 26,
A 3 (x) = 3(1 − x) 2K + 2C + γ 2 − 5 − 6( K + C + αγ + β − 2), A 4 (x) = n(n + 2)(n + γ + 2c )(n + γ + 2c − 2), where
K = (n + c )(n + γ + c ),
C = (c − 1)(c + α + β). (−1/2)
In the setting of the polynomials qn (x) = P −4,2n+4 (x) = C n differential equation we arrived at takes the form
2
(x, 3/2) for n −1, the fourth order
+ 10x x2 − 1 qn − x2 2n2 + 4n − 23 − 2n2 − 4n + 11 qn − 3 · x 2n2 + 4n − 3 qn + n2 (n + 2)2 qn = 0. (i v )
x2 − 1 qn
For n = 2 we get q2 (x) =
32x2 −5 , 35
and plugging this into the equation above we have
− −7x2 − 5 64 − 3 · x · 13 · 64x + 4 · 16 32x2 − 5 = 0. In Ismail’s notation we have c = 3/2,
ν = −1/2, α = β = γ = −1,
K = (n + 3/2)(n + 1/2),
C = −1/4
and
A 2 (x) = − 2n2 + 4n − 23 x2 − 52x + 2n2 + 4n + 3,
A 3 (x) = −3 2n2 + 4n − 3 x, A 4 (x) = n2 (n + 2)2 .
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
2865
When n = 2 we see that A 2 (x) is slightly off from our corresponding coefficient and q2 (x) = 32x35−5 does not satisfy the above equation. So there seems to be a typo in the formulation of A 2 (x) given in [13] and the families of fourth order differential equations (in the variable x) given in the book by Ismail for the associated ultraspherical polynomials are different from the families of fourth order linear differential equations (in the variable c) derived in our work above, i.e. (3.22) and (3.24). 2
5. Orthogonality of P −2,n Polynomials P −2,n satisfy the same recurrence relation as polynomials P −4,n but have different initial conditions. Hence, we do know immediately whether they are associated ultraspherical polynomials. Here we will give an independent proof of the orthogonality of these polynomials. For q s (c ) = P −2,2s (c ) we have
q−1 = P −2,−2 = 1,
q0 = P −2,0 = 0,
q2 = P −2,4 = 8c /35,
q1 = P −2,2 = 1/5,
q3 = P −2,6 =
32c 2 − 7 105
.
If q s (c ) = P −2,2s (c ) we want polynomials with index n giving the degree of the polynomial, then we will set
q¯ n := qn+1 ,
n −1,
while we ignore the “first” two polynomials; q−1 = 1 and q0 = 0. Then
q¯ 0 = 1/5,
q¯ 1 = 8c /35,
q¯ 2 =
32c 2 − 7 105
,
q¯ 3 =
8c (64c 2 − 29) 1155
(5.1)
and (4.2), becomes
4(n + 2)cqn+1 = (2n + 7)qn+2 + (2n + 1)qn ,
(5.2)
4(n + 2)cq¯ n = (2n + 7)¯qn+1 + (2n + 1)¯qn−1 ,
(5.3)
or
where the polynomials q¯ n are of degree n in c. We recall the following result by Favard which can be found in [8] or in Theorem 4.4 of [5]: Theorem 5.0.3. Let {φn , n 0} be a sequence of polynomials (monic) where pn is a polynomial of degree n, satisfying the following recursion
φn = (x − μn )φn−1 − λn φn−2 ,
n = 1, 2, 3, . . . ,
(5.4)
for some complex numbers μn , λn , n = 1, 2, 3, . . . and φ−1 = 0, φ0 = 1. Then {φn , n 0} is a sequence of orthogonal polynomials with respect to a (unique) weight measure if and only if μn ∈ R and λn+1 > 0 for all n 1. We have the following equivalent version of the theorem above.
2866
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
Theorem 5.0.4. Let { pn , n 0} be a sequence of polynomials where pn is a polynomial of degree n, satisfying the following recursion
xpn = an+1 pn+1 + bn pn + cn−1 pn−1 ,
n = 0, 1, 2, . . . ,
(5.5)
for some complex numbers an+1 , bn , cn−1 , n = 0, 1, 2, . . . and p −1 = 0. Then { pn , n 0} is a sequence of orthonormal polynomials with respect to some (unique) weight measure if and only if bn ∈ R and cn = a¯ n+1 = 0 for all n 0. Proof. The equivalence of this statement with the original result of Favard is established in the following way: Let kn = ||φn || and put pn = kn−1 φn . Then from (5.4) we get
xpn = kn−1 kn+1 pn+1 + μn+1 pn + kn−1 λn+1 kn−1 pn−1 . By putting an+1 = kn−1 kn+1 , bn = μn+1 and cn−1 = kn−1 λn+1 kn−1 we obtain (5.5) with bn ∈ R and cn = a¯ n+1 = 0 for all n 0, because
cn = (xpn+1 , pn ) = ( pn+1 , xpn ) = a¯ n+1 . Conversely, let dn be the leading coefficient of pn and put φn = dn−1 pn . Then from (5.5) we get
xφn−1 = dn−−11 an dn φn + bn−1 φn−1 + dn−−11 cn−2 dn−2 φn−2 . Since the φn ’s are monic we have dn−−11 an dn = 1. Therefore we obtain (5.4) with λn+1 = dn−1 cn−1 dn−1 = an cn−1 = an a¯ n > 0 for all n 1. 2
μn = bn−1 ∈ R and
+7 2n+1 We apply Theorem 5.0.4 to our setting. If we set an+1 = 42n (n+2) , bn = 0, cn−1 = 4(n+2) , then the hypothesis of Favard’s Theorem is not satisfied. But with a modification to our recursion formula we show that the q¯ n are orthogonal.
Theorem 5.0.5. The polynomials P −2,n (c ) are orthogonal with respect to some weight function. Proof. It is sufficient to check that there exist a family of orthonormal polynomials f n and constants λn such that qn = λn f n for all n We have the following recursion for f n :
4(n + 2)c f n = (2n + 7)λn+1 f n+1 + (2n + 1)λn−1 f n−1 or
c fn =
(2n + 7)λn+1 (2n + 1)λn−1 f n +1 + f n −1 , 4(n + 2)λn 4(n + 2)λn
for n 1. Set
A n +1 =
(2n + 7)λn+1 , 4(n + 2)λn
C n −1 =
(2n + 1)λn−1 . 4(n + 2)λn
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
(2n+5)λ
Then A n = C n−1 if and only if 4(n+1)λ n = n −1
λn2 =
2867
(2n+1)λn−1 or 4(n+2)λn
(n + 1)(2n + 1) 2 λ . (n + 2)(2n + 5) n−1
Taking λ0 = 1 we can find a family of constants λn satisfying this relation. Hence, by Theorem 5.0.4 the polynomials f n form an orthonormal family with respect to some measure, and therefore the polynomials are orthogonal. 2 One can also show that P −2,n are non-classical orthogonal polynomials. We will give a simple proof of this fact in Appendix A. The same argument works in the case of P −4,n . Acknowledgments The second author was supported in part by the CNPq grant (301743/2007-0), by the Fapesp grant (2010/50347-9) and by the MathAmSud grant (611/2012). Appendix A Given a sequence of orthogonal polynomials {qn }n0 we consider the complex linear space of all differential operators with complex coefficients on the real line that have the polynomials qn as their eigenfunctions. Thus
D( w ) = D: Dqn = γn ( D )qn , γn ( D ) ∈ C for all n 0 . Some properties of D ( w ), see [12]: (1) The definition of D ( w ) depends only on the weight function w = w (x) and not on the orthogonal sequence {qn }n0 . (2) If D ∈ D ( w ), then
D=
s
i d f i (x) , dx
i =0
where f i (x) is a polynomial and deg f i i. To ease the notation if ν ∈ C let
[ν ]i = ν (ν − 1) · · · (ν − i + 1), s
[ν ]0 = 1.
(3) If {qn }n0 is a sequence of orthogonal polynomials and D ∈ D ( w ) is of the form D =
i =0
d i f i (x)( dx ) , with f i (x) =
i
j =0
f ji ( D )x j , then
γn ( D ) =
s [n]i f ii ( D ). i =0
Let us consider the Weyl algebra D = { D =
D=
D=
s i =0
s
i =0
d i f i (x)( dx ) : f i ∈ C[x]}, and the subalgebra
i d f i (x) ∈ D: deg( f i ) i . dx
2868
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
(4) If D ∈ D satisfies the symmetry condition ( Dp , q) = ( p , Dq) for all p , q ∈ C[x], then D ∈ D ( w ). (5) For any D ∈ D ( w ) there is a unique D ∗ ∈ D ( w ) such that ( Dp , q) = ( p , D ∗ q) for all p , q ∈ C[x]. We refer to D ∗ as the adjoint of D. The map D → D ∗ is a ∗-operation in the algebra D ( w ), and the orders of D and D ∗ coincide. (6) The set S ( w ) = { D ∈ D ( w ): D = D ∗ } of all symmetric differential operators is a real form of D ( w ):
D( w ) = S ( w ) ⊕ i S ( w ). (7) D ∈ D ( w ) is symmetric if and only if γn ( D ) is real for all n 0. In the literature a weight w is called classical if there exists a second order symmetric differential operator D such that Dqn = γn qn for all n 0. Theorem A.0.6. The sequences of orthogonal polynomials (5.1) {¯qn }n0 and (4.3) {qn }n0 are not classical. Proof. In the polynomials above we replace c by x as it is more natural to use this later letter as our variable. From (6) above, to prove the theorem is equivalent to prove that in D ( w ) there is no differential operator of order 2. Suppose D ∈ D ( w ) is of order 2. Then we can assume that D is of the form
D = ax2 + bx + c
2
d
+ (ex + f )
dx
d dx
(A.1)
,
for some a, b, c , e , f ∈ C. For (5.1) the first five terms of the orthogonal sequence {¯qn }n0 are:
q¯ 0 =
1 5
q¯ 1 =
,
q¯ 4 =
8x 35
q¯ 2 =
,
32x2 − 7 105
,
q¯ 3 =
8x(64x2 − 29) 1155
,
(160 × 64)x4 − (32 × 222)x2 + (77 × 7) . 13 × 1155
From the definition of D ( w ) and (3) we know that
ax2 + bx + c
2
d dx
q¯ n + (ex + f )
d dx
q¯ n = an(n − 1) + bn + c + en + f q¯ n ,
(A.2)
for all n 0. If we set n = 0 we get c + f = 0. If we put n = 1 in the above equation we get
8
(ex + f )
35
= (b + c + e + f )
8x 35
,
which implies f = 0 and so c = 0 (as c + f = 0) and e = b + e. Hence b = 0. Then for n = 2 we obtain
(2a + 2e )
32x2 − 7 105
2
= ax
= ax2
d
2
32x2 − 7
dx 64 105
105
+ ex
+ ex
d dx
32x2 − 7
105
64x 105
and hence a + e = 0. Using all this information, the above Eq. (A.2) for n = 3 is equivalent to
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
(6a + 3e )
8x(64x2 − 29) 1155
= ax2 =
d
2
8x(64x2 − 29)
dx
1155
8 · 64 · 6 · ax3 1155
+
+ ex
8 · 64 · 3 · ex3 1155
−
d
2869
8x(64x2 − 29)
dx
8 · 29ex 1155
1155
,
which reduces to
2a + e = 0. Therefore a = e = 0 which implies that D = 0. In other words we have proved that the only differential operators in D ( w ) of degree less or equal to two are the constants. For (4.3)
q 0 = 1,
q1 = 4x/5,
A similar analysis yields also D = 0.
q2 =
32x2 − 5 35
,
q3 =
16 2 x 8x − 3 .
105
2
References [1] George B. Arfken, Hans J. Weber, Mathematical Methods for Physicists, fifth edition, Harcourt/Academic Press, Burlington, MA, 2001. [2] Joaquin Bustoz, Mourad E.H. Ismail, The associated ultraspherical polynomials and their q-analogues, Canad. J. Math. 34 (3) (1982) 718–736. [3] Murray Bremner, Universal central extensions of elliptic affine Lie algebras, J. Math. Phys. 35 (12) (1994) 6685–6692. [4] Ben Cox, Vyacheslav Futorny, DJKM algebras I: their universal central extension, Proc. Amer. Math. Soc. 139 (10) (2011) 3451–3460. [5] T.S. Chihara, An Introduction to Orthogonal Polynomials, Dover Publishers, New York, 2011. [6] Etsuro¯ Date, Michio Jimbo, Masaki Kashiwara, Tetsuji Miwa, Landau–Lifshitz equation: solitons, quasiperiodic solutions and infinite-dimensional Lie algebras, J. Phys. A 16 (2) (1983) 221–236. [7] Pavel I. Etingof, Igor B. Frenkel, Alexander A. Kirillov Jr., Lectures on Representation Theory and Knizhnik–Zamolodchikov Equations, Math. Surveys Monogr., vol. 58, Amer. Math. Soc., Providence, RI, 1998. [8] J. Favard, Sur les polynomes de Tchebicheff, C. R. Acad. Sci. Paris 200 (1935) 2052–2053. [9] Boris Feigin, Edward Frenkel, Integrable hierarchies and Wakimoto modules, in: Differential Topology, Infinite-Dimensional Lie Algebras, and Applications, in: Amer. Math. Soc. Transl. Ser. 2, vol. 194, Amer. Math. Soc., Providence, RI, 1999, pp. 27–60. [10] Edward Frenkel, Wakimoto modules, opers and the center at the critical level, Adv. Math. 195 (2) (2005) 297–404. [11] Edward Frenkel, Langlands Correspondence for Loop Groups, Cambridge Stud. Adv. Math., vol. 103, Cambridge University Press, Cambridge, 2007. [12] F. Alberto Grünbaum, Juan Tirao, The algebra of differential operators associated to a weight matrix, Integral Equations Operator Theory 58 (4) (2007) 449–475. [13] Mourad E.H. Ismail, Classical and Quantum Orthogonal Polynomials in One Variable, Encyclopedia Math. Appl., vol. 98, Cambridge University Press, Cambridge, 2005, with two chapters by Walter Van Assche, with a foreword by Richard A. Askey. [14] Igor Moiseevich Krichever, S.P. Novikov, Algebras of Virasoro type, Riemann surfaces and strings in Minkowski space, Funktsional. Anal. i Prilozhen. 21 (4) (1987) 47–61, 96. [15] Igor Moiseevich Krichever, S.P. Novikov, Algebras of Virasoro type, Riemann surfaces and the structures of soliton theory, Funktsional. Anal. i Prilozhen. 21 (2) (1987) 46–63. [16] Igor Moiseevich Krichever, S.P. Novikov, Algebras of Virasoro type, the energy-momentum tensor, and operator expansions on Riemann surfaces, Funktsional. Anal. i Prilozhen. 23 (1) (1989) 24–40. [17] H.L. Krall, On Orthogonal Polynomials Satisfying a Certain Fourth Order Differential Equation, Pennsylvania State College Studies, vol. 6, 1940, p. 24. [18] Allan M. Krall, Orthogonal polynomials satisfying fourth order differential equations, Proc. Roy. Soc. Edinburgh Sect. A 87 (3–4) (1980/1981) 271–288. [19] M. Kaneko, D. Zagier, Supersingular j-invariants, hypergeometric series, and Atkin’s orthogonal polynomials, in: Computational Perspectives on Number Theory, Chicago, IL, 1995, in: AMS/IP Stud. Adv. Math., vol. 7, Amer. Math. Soc., Providence, RI, 1998, pp. 97–126. [20] Lance L. Littlejohn, Samuel D. Shore, Nonclassical orthogonal polynomials as solutions to second order differential equations, Canad. Math. Bull. 25 (3) (1982) 291–295.
2870
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
[21] Martin Schlichenmaier, Higher genus affine algebras of Krichever–Novikov type, Mosc. Math. J. 3 (4) (2003) 1395–1427. [22] Martin Schlichenmaier, Local cocycles and central extensions for multipoint algebras of Krichever–Novikov type, J. Reine Angew. Math. 559 (2003) 53–94. g ,2 and sl g ,2 , in: Fundamental Mathe[23] O.K. Sheinman, Second-order Casimirs for the affine Krichever–Novikov algebras gl matics Today, Nezavis. Mosk. Univ., Moscow, 2003, pp. 372–404 (in Russian). [24] O.K. Sheinman, Highest-weight representations of Krichever–Novikov algebras and integrable systems, Uspekhi Mat. Nauk 60 (2(362)) (2005) 177–178. [25] M. Schlichenmaier, O.K. Scheinman, The Sugawara construction and Casimir operators for Krichever–Novikov algebras, in: Complex Analysis and Representation Theory, vol. 1, J. Math. Sci. (N. Y.) 92 (2) (1998) 3807–3834. [26] M. Shlikhenmaier, O.K. Sheinman, The Wess–Zumino–Witten–Novikov theory, Knizhnik–Zamolodchikov equations, and Krichever–Novikov algebras, Uspekhi Mat. Nauk 54 (1(325)) (1999) 213–250. [27] V.V. Schechtman, A.N. Varchenko, Hypergeometric solutions of Knizhnik–Zamolodchikov equations, Lett. Math. Phys. 20 (4) (1990) 279–283. (1) [28] Minoru Wakimoto, Fock representations of the affine Lie algebra A 1 , Comm. Math. Phys. 104 (4) (1986) 605–609. [29] G.N. Watson, A Treatise on the Theory of Bessel Functions, Cambridge Math. Lib., Cambridge University Press, Cambridge, 1995, reprint of the second (1944) edition. [30] Jet Wimp, Explicit formulas for the associated Jacobi polynomials and some applications, Canad. J. Math. 39 (4) (1987) 983–1000.
Contents lists available at SciVerse ScienceDirect
Journal of Differential Equations www.elsevier.com/locate/jde
DJKM algebras and non-classical orthogonal polynomials Ben Cox a,∗ , Vyacheslav Futorny b , Juan A. Tirao c a b c
Department of Mathematics, College of Charleston, 66 George St., Charleston, SC 29424, USA Department of Mathematics, University of São Paulo, São Paulo, Brazil CIEM-FAMAF, Universidad Nacional de Cordoba, 5000 Cordoba, Argentina
a r t i c l e
i n f o
Article history: Received 27 January 2013 Revised 11 June 2013 Available online 2 August 2013
a b s t r a c t We describe families of polynomials arising in the study of the universal central extensions of Lie algebras introduced by Date, Jimbo, Kashiwara, and Miwa (1983) [6] in their work on the Landau– Lifshitz equations. We show these two families of polynomials satisfy certain fourth order linear differential equations by direct computation and one of the families is a particular collection of associated ultraspherical polynomials. © 2013 Elsevier Inc. All rights reserved.
1. Introduction Date, Jimbo, Kashiwara and Miwa [6] studied integrable systems arising from Landau–Lifshitz differential equation. The hierarchy of this equation is written in terms of free fermions on an elliptic curve. The authors introduced an infinite dimensional Lie algebra which is a one dimensional central extension of g ⊗ C[t , t −1 , u | u 2 = (t 2 − b2 )(t 2 − c 2 )] where b = ±c are complex constants and g is a simple finite dimensional Lie algebra. This Lie algebra which we call the DJKM algebra, acts on the solutions of the Landau–Lifshitz equation as infinitesimal Bäcklund transformations. The Lie algebra above is an example of a Krichever–Novikov algebra (see [15,14,16]). A fair amount of interesting and fundamental work has be done by Krichever, Novikov, Schlichenmaier, and Sheinman on the representation theory of these algebras. In particular Wess–Zumino–Witten–Novikov theory and analogues of the Knizhnik–Zamolodchikov equations are developed for these algebras (see the survey article [24], and for example [26,23,21,22,25]). In [4] the authors gave commutations relations of the universal central extension of the DJKM Lie algebra in terms of a basis of the algebra and certain polynomials. More precisely in order to pin
*
Corresponding author. E-mail addresses: [email protected] (B. Cox), [email protected] (V. Futorny), [email protected] (J.A. Tirao).
0022-0396/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jde.2013.07.020
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
2847
down this central extension, we needed to describe four families of polynomials that appeared as coefficients in the commutator formulae. In this previous work we gave recursion relations for these polynomials and then found generating functions for them. Two of these families of polynomials are given in terms of elliptic integrals and the other two families are slight variations of ultraspherical polynomials. The main purpose of this note is to describe fourth order linear differential equations satisfied by the these two elliptic families of polynomials (see (3.22) and (3.24)), and to explain why these polynomials are orthogonal and non-classical (see Corollary 4.0.2, Theorem 5.0.5, and Theorem A.0.6). In fact one of the families are particular examples of associated ultraspherical polynomials (see [2]). The associated ultraspherical polynomials in turn are, up to factors of ascending factorials, particular associated Jacobi polynomials. The associated Jacobi polynomials are known to satisfy certain fourth order linear differential equations (see [13] and formula (48) in [30]). The proof given in the previous cited paper is obtained using MACSYMA and ideas coming from the classic text [29]. We show how we obtained a proof in the ultraspherical case that is done by hand and omit the calculation done for the remaining family of polynomials as it is similar. The referee was instrumental in identifying one of the families of non-classical polynomials describing the universal central extension and the references about associated ultraspherical and Jacobi polynomials. We would like to thank the referee for directing us to the relevant articles and books. For other examples of families of orthogonal polynomials that satisfy fourth order linear differential equations see the work of Samuel Shore, Allan Krall, and H.L. Krall [20,18,17]. The authors would like to thank Lance Littlejohn for these references and helpful correspondence. Still for even more examples of families of orthogonal polynomials that satisfy fourth order linear differential equations appearing in the study of supersingular j-invariants see the work of Kaneko and Zagier [19]. We plan to use these families of polynomials to describe free field realizations of the DJKM algebra in the setting of conformal field theory. In the case of affine Kac–Moody algebras, initial motivation for the use of Wakimoto’s realization (or free field realization) was to prove a conjecture of V. Kac and D. Kazhdan on the character of certain irreducible representations of affine Kac–Moody algebras at the critical level (see [28] and [10]). Another motivation for constructing free field realizations is that they are used to provide integral solutions to the KZ-equations (see for example [27] and [7] and their references). A third is that they are used to help in determining the center of a certain completion of the enveloping algebra of an affine Lie algebra at the critical level which is an important ingredient in the geometric Langland’s correspondence [11]. Yet a fourth is that free field realizations of an affine Lie algebra appear naturally in the context of the generalized AKNS hierarchies [9]. 2. DJKM algebras Let R be a commutative algebra defined over C. Consider the left R-module with action f ( g ⊗ h) = f g ⊗ h for f , g , h ∈ R and let K be the submodule generated by the elements 1 ⊗ f g − f ⊗ g − g ⊗ f . Then Ω R1 = F / K is the module of Kähler differentials. The element f ⊗ g + K is traditionally denoted by f dg. The canonical map d : R → Ω R1 by df = 1 ⊗ f + K . The exact differentials are the elements of the subspace dR. The coset of f dg modulo dR is denoted by f dg. As C. Kassel showed the universal central extension of the current algebra g ⊗ R where g is a simple finite dimensional Lie algebra defined over C, is the vector space g = (g ⊗ R ) ⊕ Ω R1 /dR with Lie bracket given by
[x ⊗ f , Y ⊗ g ] = [xy ] ⊗ f g + (x, y ) f dg ,
[ x ⊗ f , ω ] = 0,
ω , ω = 0,
where x, y ∈ g, and ω, ω ∈ Ω R1 /dR and (x, y ) denotes the Killing form on g. Consider the polynomial
p (t ) = t n + an−1 t n−1 + · · · + a0 , where ai ∈ C and an = 1. Fundamental to the description of the universal central extension for R = C[t , t −1 , u | u 2 = p (t )] are the following two results:
2848
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
Theorem 2.0.1. (See [3], Theorem 3.4.) Let R be as above. The set
t −1 dt , t −1 u dt , . . . , t −n u dt
forms a basis of Ω R1 /dR (omitting t −n u dt if a0 = 0). Proposition 2.0.2. (See [4], Lemma 2.0.2.) If um = p (t ) and R = C[t , t −1 , u | um = p (t )], then in Ω R1 /dR, one has n −1 (m + 1)n + im t n+i −1 u dt ≡ − (m + 1) j + mi a j t i + j −1 u dt mod dR .
(2.1)
j =0
In the Date–Jimbo–Miwa–Kashiwara setting one takes m = 2 and p (t ) = (t 2 − a2 )(t 2 − b2 ) = t 4 − (a2 + b2 )t 2 + (ab)2 with a = ±b and neither a nor b is zero. We fix from here onward R = C[t , t −1 , u | u 2 = (t 2 − a2 )(t 2 − b2 )]. As in this case a0 = (ab)2 , a1 = 0, a2 = −(a2 + b2 ), a3 = 0 and a4 = 1, then letting k = i + 3 the recursion relation in (2.1) looks like
(6 + 2k)t k u dt = −2(k − 3)(ab)2 t k−4 u dt + 2k a2 + b2 t k−2 u dt . √
After a change of variables, u → u /ab, t → t / ab, we may assume that a2 b2 = 1. Then the recursion relation looks like
(6 + 2k)t k u dt = −2(k − 3)t k−4 u dt + 4kct k−2 u dt ,
(2.2)
after setting c = (a2 + b2 )/2, so that p (t ) = t 4 − 2ct 2 + 1. Let P k := P k (c ) be the polynomial in c satisfy the recursion relation
(6 + 2k) P k (c ) = 4kc P k−2 (c ) − 2(k − 3) P k−4 (c ) for k 0. Then set
P (c , z) :=
P k (c ) zk+4 =
k−4
P k−4 (c ) zk ,
k0
so that after some straightforward rearrangement of terms we have
0=
(6 + 2k) P k (c ) zk − 4c
k0
k P k−2 (c ) zk + 2
k0
(k − 3) P k−4 (c ) zk
k0
d = −2z−4 + 8cz−2 − 6 P (c , z) + 2z−3 − 4cz−1 + 2z P (c , z)
+ 2z
−4
− 8cz
−2
dz
P −4 (c ) − 4c P −3 (c ) z
−1
− 2P −2 (c ) z−2 − 4P −1 (c ) z−1 .
Hence P (c , z) must satisfy the differential equation
d dz
P (c , z) −
3z4 − 4cz2 + 1 z5
− 2cz3
+z
P (c , z) =
2( P −1 + c P −3 ) z3 + P −2 z2 + (4cz2 − 1) P −4 z5 − 2cz3 + z
.
(2.3)
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
2849
This has integrating factor
μ(z) = exp
1 −2( z3 − cz) − dz z 1 − 2cz2 + z4
1 = exp − ln 1 − 2cz2 + z4 − ln( z) = √
1
z 1 − 2cz2 + z4
2
.
2.1. Elliptic case 1 If we take initial conditions P −3 (c ) = P −2 (c ) = P −1 (c ) = 0 and P −4 (c ) = 1 then we arrive at a generating function
P −4 (c , z) :=
P −4,k (c ) zk+4 =
k−4
P −4,k−4 (c ) zk
k0
defined in terms of an elliptic integral
P −4 (c , z) = z 1 − 2cz2 + z4
4cz2 − 1 z2 ( z4 − 2cz2 + 1)3/2
dz.
2.2. Elliptic case 2 If we take initial conditions P −4 (c ) = P −3 (c ) = P −1 (c ) = 0 and P −2 (c ) = 1, we arrive at a generating function defined in terms of another elliptic integral:
P −2 (c , z) = z 1 − 2cz2 + z4
1
( z4 − 2cz2 + 1)3/2
dz.
2.3. Gegenbauer case 3 If we take P −1 (c ) = 1, and P −2 (c ) = P −3 (c ) = P −4 (c ) = 0 and set P −1 (c , z) = then
P −1 (c , z) = (−1/2)
where Q n
1
3
c2 − 1
2 3
cz − z − cz + c z −
∞
n0
P −1,n−4 zn ,
(−1/2)
cQn
(c ) z
2n+1
,
k =2
(c ) is the n-th Gegenbauer polynomial. Hence P −1,−4 (c ) = P −1,−3 (c ) = P −1,−2 (c ) = P −1,2m (c ) = 0, P −1,−1 (c ) = 1, P −1,2n−3 (c ) = (−1/2)
for m 0 and n 2. The Q n
1 − c2
−c Q n (c ) , c2 − 1
(c ) are known to satisfy the second order differential equation:
d2 d2 c
(−1/2)
Qn
(−1/2)
(c ) + n(n − 1) Q n
(c ) = 0,
so that the P −1,k := P −1,k (c ) satisfy the second order differential equation
2850
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
c4 − c2
d2 d2 c
P −1,2n−3 + 2c c 2 + 1
d dc
P −1,2n−3 + −c 2 n(n − 1) − 2 P −1,2n−3 = 0
for n 2. 2.4. Gegenbauer case 4 Next we consider the initial conditions P −1 (c ) = 0 = P −2 (c ) = P −4 (c ) = 0 with P −3 (c ) = 1 and set
P −3 (c , z) =
P −3,n−4 (c ) z =
n0
(−1/2)
where Q n
1
n
2
c2 − 1
3
3
c z − cz − z + cz −
∞
(−1/2)
Qn
(c ) z
2n+1
,
k =2
(c ) is the n-th Gegenbauer polynomial. Hence P −3,−4 (c ) = P −3,−2 (c ) = P −3,−1 (c ) = P −1,2m (c ) = 0, P −3,−3 (c ) = 1, P −3,2n−3 (c ) =
− Q n (c ) , c2 − 1
for m 0 and n 2 and hence
c2 − 1
d2 d2 c
P −3,2n−3 + 4c
d dc
P −3,2n−3 − (n + 1)(n − 2) P −3,2n−3 = 0
for n 2 and P −1,2n−3 = c P −3,2n−3 for n 2. We’ll see in the last section how the polynomials described in the elliptic case 1 are particular examples of associated ultraspherical polynomials. The importance of these families of polynomials come from our previous work describing the universal central extension of the DJKM algebra: Theorem 2.4.1. (See [4].) Let g be a simple finite dimensional Lie algebra over the complex numbers with the Killing form ( | ) and define ψi j (c ) ∈ Ω R1 /dR by
ψi j (c ) =
⎧ω i + j −2 ⎪ ⎪ ⎪ ⎨ P −3,i + j −2 (c )(ω−3 + c ω−1 )
for i + j = 1, 0, −1, −2,
for i + j = 2n − 1 3, n ∈ Z, ⎪ P ( c )( c ω + ω ) for i + j = −2n + 1 −3, n ∈ Z, −3 −1 −3 , i + j −2 ⎪ ⎪ ⎩ P −4,|i + j |−2 (c )ω−4 + P −2,|i + j |−2 (c )ω−2 for |i + j | = 2n 2, n ∈ Z.
(2.4)
The universal central extension of the Date–Jimbo–Kashiwara–Miwa algebra is the Z2 -graded Lie algebra
g = g0 ⊕ g1 , where
g0 = g ⊗ C t , t −1 ⊕ Cω0 , with bracket
g1 = g ⊗ C t , t −1 u ⊕ Cω−4 ⊕ Cω−3 ⊕ Cω−2 ⊕ Cω−1
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
x ⊗ t i , y ⊗ t j = [x, y ] ⊗ t i + j + δi + j ,0 j (x, y )ω0 ,
x ⊗ t i −1 u , y ⊗ t j −1 u = [x, y ] ⊗ t i + j +2 − 2ct i + j + t i + j −2
2851
+ δi + j ,−2 ( j + 1) − 2c j δi + j ,0 + ( j − 1)δi + j ,2 (x, y )ω0 ,
x ⊗ t i −1 u , y ⊗ t j = [x, y ]u ⊗ t i + j −1 + j (x, y )ψi j (c ).
3. Differential equations for elliptic type 1 and 2 3.1. Elliptic type 1 How we arrive at the fourth order linear differential equation that the polynomials P −4,n satisfy stems from the approach used in Afken’s book [1] for finding the second order linear differential equation that Legendre polynomials satisfy using only the recursion relation they satisfy. His technique seems to be indirectly based on ideas used in the theory of Gröbner basis. Unfortunately the calculations and relations involved are rather tedious. From now on we are going to reindex the polynomials P −4,n :
P −4 (c , z) = z 1 − 2cz2 + z4
4cz2 − 1
dz = z2 ( z4 − 2cz2 + 1)3/2 1
∞
P −4,n (c ) zn
n =0
16 2 = 1 + z4 + z6 + 32c 2 − 5 z8 + c 8c − 3 z10 5 35 105 4c
−
(2048c 4 − 1248c 2 + 75) 1155
z12 + O z14 .
This means that now P −4,0 (c ) = 1, P −4,1 (c ) = P −4,2 (c ) = P −4.3 (c ) = 0. Besides P −4,0 (c ), the first few nonzero polynomials in c are
P −4,4 (c ) = 1, P −4,10 =
P −4,6 =
16 2 c 8c − 3 , 105
4c 5
,
P −4,12 = −
P −4,8 =
32c 2 − 5 35
,
(2048c 4 − 1248c 2 + 75) 1155
and P −4,n (c ) satisfy the following recursion:
(6 + 2k) P k+4 (c ) = 4kc P k+2 (c ) − 2(k − 3) P k (c ).
(3.1)
Our goal in this section is to find families of linear differential equation in c that these polynomials satisfy. We start off with the generating function
P −4 (c , z) = z 1 − 2cz2 + z4
=z
1 − 2cz2
+
z4
4cz2 − 1 z2 ( z4 − 2cz2 + 1)3/2 ∞ (3/2) 4c Q n (c ) n =0
(λ)
2n + 1
z
2n+1
dz
∞ (3/2) Qn (c ) 2n−1 − z , 2n − 1 n =0
where Q n (c ) is the n-Gegenbauer polynomial. These polynomials satisfy the second order linear ODE:
2852
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
1 − c 2 y − (2λ + 1)c y + n(n + 2λ) y = 0,
where the derivative is with respect to c. Thus for λ = 3/2 we get
1 − c2
(3/2)
Qn
(3/2) (3/2) (c ) − 4c Q n (c ) + n(n + 3) Q n (c ) = 0.
(3.2)
Rewrite the expansion formula for P −4 (c , z) to get
z−1 1 − 2cz2 + z4
−1/2
P −4 (c , z) =
∞ (3/2) 4c Q n (c )
2n + 1
n =0
z2n+1 −
∞ (3/2) Qn (c ) 2n−1 , z 2n − 1
(3.3)
n =0
2
d d and apply the differential operator L := (1 − c 2 ) dc 2 − 4c dc to the right hand side to get
(3/2)
L 4c Q n
d2 d (3/2) (c ) = 1 − c 2 − 4c 4c Q n (c ) 2 dc
dc
(3/2) = −8(n + 1) Q n+1 (c ) − 4c n2
(3/2) + n − 2 Q n (c )
using the identity
1 − c2
d dc
(λ)
(λ)
(λ)
Q n (c ) = (n + 2λ)c Q n (c ) − (n + 1) Q n+1 (c ).
(3.4)
Using a number of simplifications that we have put on http://arxiv.org/archive/math we get
L
∞ (3/2) 4c Q n (c ) n =0
2n + 1
z
2n+1
=
(3/2) (3/2) ∞ −8(n + 1) Q n+1 (c ) − 4c (n2 + n − 2) Q n (c )
2n + 1
n =0
= −4
1 z( z4 − 2cz2 + 1)3/2
− cz2
dz
+ 9c
d
1
−4
z2 ( z4 − 2cz2 + 1)3/2
1
( z4 − 2cz2 + 1)3/2 1
( z4 − 2cz2 + 1)3/2
z2n+1
−
dz
cz
( z4 − 2cz2 + 1)3/2
dz.
In addition we have
∞ ∞ (3/2) (3/2) Qn (c ) 2n−1 n(n + 3) Q n (c ) 2n−1 L z z =− 2n − 1 2n − 1 n =0
n =0
∞ (3/2) 1 d2 9 d 7 Qn (c ) 2n−1 = − z2 2 + z + z 4 dz 4 dz 4 2n − 1 n =0 1 d 1 9 − = − z2 4 dz z2 ( z4 − 2cz2 + 1)3/2 4z( z4 − 2cz2 + 1)3/2
7 1 − dz. 4 z2 ( z4 − 2cz2 + 1)3/2
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
2853
Thus the right hand side of (3.3) becomes
−4
1 z( z4 − 2cz2 + 1)3/2
d
− cz2
dz
=− − +
1
−
( z4 − 2cz2 + 1)3/2
( z4 − 2cz2 + 1)3/2
d
1
4
dz
z2 ( z4 − 2cz2 + 1)3/2
7
4
1 z2 ( z4 − 2cz2 + 1)3/2 7
4z( z4 − 2cz2 + 1)3/2 cz
( z4 − 2cz2 + 1)3/2 9
1
√
4 z z4 − 2cz2 + 1
+
−
cz
( z4 − 2cz2 + 1)3/2
dz
1
+ z2
dz
z2 ( z4 − 2cz2 + 1)3/2
1
+ 9c
+
1
−4
+
9 4z( z4 − 2cz2 + 1)3/2
dz 6cz2 ( z2 − c )
( z4 − 2cz2 + 1)5/2 2z2 (4z4 − 5cz2 + 1)
z3 (−2cz2 + z4 + 1)5/2
P −4 (c , z).
Applying the differential operator L to the left hand side of (3.3), we get
L z−1 1 − 2cz2 + z4
−1/2
P −4 (c , z)
d 2 −1 −1/2 −1/2 d −1 z 1 − 2cz2 + z4 P −4 (c , z) − 4c z 1 − 2cz2 + z4 P −4 (c , z) = 1 − c2 2 dc
dc
=
z(−4c + (3 + 5c ) z − 4cz ) 2
2
(1 − 2cz2 + z4 )5/2 +
4
P −4 (c , z) +
−4c + (2 + 6c 2 ) z2 − 4cz4 d P −4 (c , z) dc z(1 − 2cz2 + z4 )3/2
(1 − c 2 ) d2 P −4 (c , z) . z(1 − 2cz2 + z4 )1/2 dc 2
Hence we have
z(−4c + (3 + 5c 2 ) z2 − 4cz4 )
(1 − 2cz2
+
z4 )5/2
P −4 (c , z) +
−4c + (2 + 6c 2 ) z2 − 4cz4 d P −4 (c , z) 2 4 3 / 2 dc z(1 − 2cz + z )
(1 − c 2 ) d2 P −4 (c , z) z(1 − 2cz2 + z4 )1/2 dc 2 −4cz2 − 7 1 2 d = − cz dz ( z4 − 2cz2 + 1)3/2 4z( z4 − 2cz2 + 1)3/2 1 1 d 1 9 + c √ + z2 P −4 (c , z) 2 4 2 3 / 2 4 4 dz z ( z − 2cz + 1) 4 z z − 2cz2 + 1 +
as
−3/2 −2(4z4 − 5cz2 + 1) d −2 4 = 3 4 z z − 2cz2 + 1 . dz z ( z − 2cz2 + 1)5/2
2854
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
As a consequence
z(−4c + (3 + 5c 2 ) z2 − 4cz4 )
(1 − 2cz2
+
z4 )5/2
P −4 (c , z) +
−4c + (2 + 6c 2 ) z2 − 4cz4 d P −4 (c , z) 2 4 3 / 2 dc z(1 − 2cz + z )
d2 (1 − c 2 ) P −4 (c , z) z(1 − 2cz2 + z4 )1/2 dc 2 4cz2 + 7 6z3 (c − z2 ) − c =− 4z( z4 − 2cz2 + 1)3/2 ( z4 − 2cz2 + 1)5/2 +
−
2(4z4 − 5cz2 + 1) z( z4
− 2cz2
+ 1)5/2
+
9
1
√
4 z z4 − 2cz2 + 1
P −4 (c , z),
which gives us
9
− + 5cz2 − 4
= −
9 4
15 4
+ 4c 2 z4 + 5cz6
z4 − 2cz2 + 1
2
+ z2 −4c + 3 + 5c 2 z2 − 4cz4 P −4 (c , z)
d P −4 (c , z) + −4c + 2 + 6c 2 z2 − 4cz4 z4 − 2cz2 + 1 dc
2 d P −4 (c , z) . + 1 − c 2 z4 − 2cz2 + 1 2 2
dc
Expanding this out in detail and writing P −4,k (c ) as P k we will obtain
0 = −9P n + 20c P n−2 − 16c 2 + 6 P n−4 + 20c P n−6 − 9P n−8
− 16c P n + 8 1 + 7c 2 P n −2 − 48c 1 + c 2 P n −4 + 8 1 + 7c 2 P n −6 − 16c P n −8 + 4 1 − c 2 P n − 4c P n−2 + 2 1 + 2c 2 P n−4 − 4c P n−6 + P n−8 .
(3.5)
We now differentiate with respect to c the recursion relations
(6 + 2k) P k+4 = 4kc P k+2 − 2(k − 3) P k
(3.6)
to get
(6 + 2k) P k +4 = 4k P k+2 + 4kc P k +2 − 2(k − 3) P k ,
(3.7)
(6 + 2k) P k+4
(3.8)
= 8k P k+2 + 4kc P k+2
− 2(k − 3) P k .
After setting k = n − 8 in the last equation we get
0 = −(2n − 10) P n−4 + 8(n − 8) P n −6 + 4(n − 8)c P n−6 − 2(n − 11) P n−8 . So multiplying (3.5) by (n − 11) and adding it to 2(1 − c 2 ) times the above gives us
(3.9)
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
2855
0 = (n − 11) −9P n + 20c P n−2 − 16c 2 + 6 P n−4 + 20c P n−6 − 9P n−8
− 16c P n + 8 1 + 7c 2 P n −2 − 48c 1 + c 2 P n −4 − 16c P n −8 + 4 1 − c 2 P n − 4c P n−2 + 8 c 2 (5n − 61) + (3n − 27) P n −6 − 4 1 − c 2 4c 2 (n − 11) + n − 17 P n−4 + 8(n − 14)c 1 − c 2 P n−6 .
(3.10)
Setting k = n − 8 in (3.7) we get
0 = −(2n − 10) P n −4 + 4(n − 8) P n−6 + 4(n − 8)c P n −6 − 2(n − 11) P n −8 .
(3.11)
Multiplying this equation by −8c and add to the previous equation we get
0 = (n − 11) −9P n + 20c P n−2 − 16c 2 + 6 P n−4 − 9P n−8
− 16c P n + 8 1 + 7c 2 P n −2 + 4 1 − c 2 P n − 4c P n−2 + 16c −3 c 2 + 1 (n − 11) + n − 5 P n −4 − 4 1 − c 2 4c 2 (n − 11) + n − 17 P n−4 − 12c (n − 3) P n−6 + 8 c 2 (n − 29) + 3(n − 9) P n −6 + 8(n − 14)c 1 − c 2 P n−6 .
Finally if we multiply the previous equation by 2 and add it to −9 times (3.6) we get
0 = 2(n − 11) −9P n + 20c P n−2 − 16c P n + 8 1 + 7c 2 P n −2 + 4 1 − c 2
P n − 4c P n−2
+ 6(n + 7) − 32c 2 (n − 11) P n−4 + 32c −3 c 2 + 1 (n − 11) + n − 5 P n −4 − 8 1 − c 2 4c 2 (n − 11) + n − 17 P n−4 − 60c (n − 6) P n−6 + 16 c 2 (n − 29) + 3(n − 9) P n −6 + 16(n − 14)c 1 − c 2 P n−6 .
This is an equation without the P n−8 in it. We now get rid of the P n−6 ’s in them. After setting k = n − 6 in (3.8) we get
0 = −(2n − 6) P n−2 + 8(n − 6) P n −4 + 4(n − 6)c P n−4 − 2(n − 9) P n−6 .
(3.12)
Now we multiply the previous equation by −8(n − 14)c (1 − c 2 ) and add it to n − 9 times the equation before it, we obtain
0 = 2(n − 11)(n − 9) −9P n + 20c P n−2 − 16c P n + 8 1 + 7c 2 P n −2 + 4 1 − c 2 P n
+ (n − 9) 6(n + 7) − 32c 2 (n − 11) P n−4 + 16c c 2 − 1 n2 − 23n + 156 P n−2 − 32c c 2 n2 − 20n + 129 + 4n2 − 86n + 420 P n −4 − 8 c 2 − 1 60c 2 + n2 − 26n + 153 P n−4 − 60c (n − 9)(n − 6) P n−6 + 16(n − 9) c 2 (n − 29) + 3(n − 9) P n −6 .
From (3.7) with k = n − 6 one has
0 = −(2n − 6) P n −2 + 4(n − 6) P n−4 + 4(n − 6)c P n −4 − 2(n − 9) P n −6 .
(3.13)
2856
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
We multiply this equation by 8(c 2 (n − 29) + 3(n − 9)) and add it to the previous equation to get
0 = 2(n − 11)(n − 9) −9P n + 20c P n−2 − 16c P n + 8 1 + 7c 2 P n −2 + 4 1 − c 2 P n
− 8(2n − 6) c 2 (n − 29) + 3(n − 9) P n −2 + 16c c 2 − 1 n2 − 23n + 156 P n−2 − 6 80c 2 (n − 5) − 17n2 + 242n − 801 P n−4 − 32c 15c 2 (n − 3) + n2 − 41n + 258 P n −4 − 8 c 2 − 1 60c 2 + n2 − 26n + 153 P n−4 − 60c (n − 9)(n − 6) P n−6 .
From (3.6) with k = n − 6 one has
0 = −(2n − 6) P n−2 + 4(n − 6)c P n−4 − 2(n − 9) P n−6 . We multiply this equation by −30c (n − 6) and add it to the previous equation to get
0 = 2(n − 11)(n − 9) −9P n − 16c P n + 8 1 + 7c 2 P n −2 + 4 1 − c 2 P n
+ 20c 5n2 − 67n + 252 P n−2 − 8(2n − 6) c 2 (n − 29) + 3(n − 9) P n −2 + 16c c 2 − 1 n2 − 23n + 156 P n−2 + 6 −20c 2 (n − 4)2 + 17n2 − 242n + 801 P n−4 − 32c 15c 2 (n − 3) + n2 − 41n + 258 P n −4 − 8 c 2 − 1 60c 2 + n2 − 26n + 153 P n−4 .
The above equation now does not have the index n − 6 in it. Now we want to eliminate the indices with n − 4 in them. From (3.8) with k = n − 4 one has
0 = −(2n − 2) P n + 8(n − 4) P n −2 + 4(n − 4)c P n−2 − 2(n − 7) P n−4 . We multiply this equation by −4(c 2 − 1)(60c 2 + n2 − 26n + 153) and add it to n − 7 times the previous equation to get
0 = 2(n − 11)(n − 9)(n − 7) −9P n − 16c P n
+ 20c (n − 7) 5n2 − 67n + 252 P n−2 + 6(n − 7) −20c 2 (n − 4)2 + 17n2 − 242n + 801 P n−4 − 32c (n − 7) 15c 2 (n − 3) + n2 − 41n + 258 P n −4 + 480 c 2 − 1 c 2 (n − 1) − n + 9 P n − 32 60c 4 (n − 4) + c 2 −2n3 + 45n2 − 484n + 1749 + 15 n2 − 14n + 45 P n −2 − 960c c 2 − 1 c 2 (n − 4) − n + 8 P n−2 .
From (3.7) with k = n − 4 one has
0 = −(2n − 2) P n + 4(n − 4) P n−2 + 4(n − 4)c P n −2 − 2(n − 7) P n −4 . We multiply this equation by −16c (15c 2 (n − 3) + n2 − 41n + 258) and add it to the previous equation to get
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
2857
0 = −18(n − 11)(n − 9)(n − 7) P n
+ 480c c 2 n2 − 4n + 3 − n2 + 4n + 29 P n + 480 c 2 − 1 c 2 (n − 1) − n + 9 P n + 12c −80c 2 n2 − 7n + 12 + 3n3 + 70n2 − 1049n + 2564 P n−2 − 480 2c 4 n2 − 5n + 4 − 3c 2 n2 − 8n + 7 + n2 − 14n + 45 P n −2 − 960c c 2 − 1 c 2 (n − 4) − n + 8 P n−2 + 6(n − 7) −20c 2 (n − 4)2 + 17n2 − 242n + 801 P n−4 .
From (3.6) with k = n − 4 one has
0 = −2(n − 1) P n + 4(n − 4)c P n−2 − 2(n − 7) P n−4 . We multiply this equation by 3(−20c 2 (n − 4)2 + 17n2 − 242n + 801) and add it to the previous equation to get
0 = 120(n − 4)2 c 2 (n − 1) − n + 9 P n
+ 480c c 2 n2 − 4n + 3 − n2 + 4n + 29 P n + 480 c 2 − 1 c 2 (n − 1) − n + 9 P n − 240c (n − 2)2 c 2 (n − 4) − n + 8 P n−2 − 480 2c 4 n2 − 5n + 4 − 3c 2 n2 − 8n + 7 + n2 − 14n + 45 P n −2 − 960c c 2 − 1 c 2 (n − 4) − n + 8 P n−2 .
This can be rewritten as
0 = (n − 4)2 c 2 (n − 1) − n + 9 P n
+ 4c c 2 n2 − 4n + 3 − n2 + 4n + 29 P n + 4 c 2 − 1 c 2 (n − 1) − n + 9 P n − 2c (n − 2)2 c 2 (n − 4) − n + 8 P n−2 − 4 2c 4 n2 − 5n + 4 − 3c 2 n2 − 8n + 7 + n2 − 14n + 45 P n −2 − 8c c 2 − 1 c 2 (n − 4) − n + 8 P n−2 .
(3.14)
We have now reduced to a differential equation with only the indices n and n − 2. There is a bit of a trick to reduce it down to a linear ODE with only the index n in it. The somewhat vague idea is to find more equations to in order to cancel all but terms with the index n in them. From (3.8) with k = n − 2 one has
0 = −2(n + 1) P n+2 + 8(n − 2) P n + 4(n − 2)c P n − 2(n − 5) P n−2 . We multiply this equation by −4c (c 2 − 1)(c 2 (n − 4) + 8 − n) and add it to n − 5 times the previous equation to get
2858
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
0 = −2(n − 5)c (n − 2)2 c 2 (n − 4) − n + 8 P n−2
− 4(n − 5) 2c 4 n2 − 5n + 4 − 3c 2 n2 − 8n + 7 + n2 − 14n + 45 P n −2 + (n − 5)(n − 4)2 c 2 (n − 1) − n + 9 P n + 4c −8c 4 n2 − 6n + 8 + c 2 n3 + 7n2 − 105n + 177 − n3 + n2 + 89n − 273 P n − 4 c 2 − 1 4c 4 n2 − 6n + 8 + c 2 −5n2 + 46n − 69 + n2 − 14n + 45 P n + 8(n + 1)c c 2 − 1 c 2 (n − 4) + 8 − n (n + 1) P n+2 .
From (3.7) with k = n − 2 one has
0 = −2(n + 1) P n +2 + 4(n − 2) P n + 4(n − 2)c P n − 2(n − 5) P n −2 . We multiply this equation by
−2 2c 4 (n − 4)(n − 1) − 3c 2 (n − 7)(n − 1) + (n − 9)(n − 5) and add it to the previous equation to get
0 = −2(n − 5)c (n − 2)2 c 2 (n − 4) − n + 8 P n−2
+ −16c 4 n3 − 7n2 + 14n − 8 + c 2 n4 + 10n3 − 171n2 + 416n − 256 − n2 n2 − 14n + 45 P n − 4c (n + 1) 4c 4 n2 − 6n + 8 + c 2 −7n2 + 60n − 93 + 3 n2 − 12n + 31 P n − 4 c 2 − 1 4c 4 n2 − 6n + 8 + c 2 −5n2 + 46n − 69 + n2 − 14n + 45 P n + 4 2c 4 (n − 4)(n − 1) − 3c 2 (n − 7)(n − 1) + (n − 9)(n − 5) (n + 1) P n +2 + 8(n + 1)c c 2 − 1 c 2 (n − 4) + 8 − n (n + 1) P n+2 .
Next we get rid of the P n−2 term by multiplying (3.6) with k = n − 2;
0 = −2(n + 1) P n+2 + 4(n − 2)c P n − 2(n − 5) P n−2 , by −c (n − 2)2 (c 2 (n − 4) − n + 8) and adding it to the previous equation:
0 = −n2 4c 4 n2 − 6n + 8 + c 2 −5n2 + 46n − 69 + n2 − 14n + 45 P n
− 4c (n + 1) 4c 4 n2 − 6n + 8 + c 2 −7n2 + 60n − 93 + 3 n2 − 12n + 31 P n − 4 c 2 − 1 4c 4 n2 − 6n + 8 + c 2 −5n2 + 46n − 69 + n2 − 14n + 45 P n + 2c (n + 1)(n − 2)2 c 2 (n − 4) − n + 8 P n+2 + 4(n + 1) 2c 4 (n − 4)(n − 1) − 3c 2 (n − 7)(n − 1) + (n − 9)(n − 5) P n +2 + 8(n + 1)2 c c 2 − 1 c 2 (n − 4) + 8 − n P n+2 .
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
2859
Letting n → n − 2 in the above equation we get
0 = −(n − 2)2 4c 4 n2 − 10n + 24 + c 2 −5n2 + 66n − 181 + n2 − 18n + 77 P n−2
− 4c (n − 1) 4c 4 n2 − 10n + 24 + c 2 −7n2 + 88n − 241 + 3 n2 − 16n + 59 P n −2 − 4 c 2 − 1 4c 4 n2 − 10n + 24 + c 2 −5n2 + 66n − 181 + n2 − 18n + 77 P n−2 + 2c (n − 1)(n − 4)2 c 2 (n − 6) − n + 10 P n + 4(n − 1) 2c 4 (n − 6)(n − 3) − 3c 2 (n − 9)(n − 3) + (n − 11)(n − 7) P n + 8(n − 1)2 c c 2 − 1 c 2 (n − 6) − n + 10 P n . (3.15)
We compare this to (3.14)
0 = −2c (n − 2)2 c 2 (n − 4) − n + 8 P n−2
− 4 2c 4 n2 − 5n + 4 − 3c 2 n2 − 8n + 7 + n2 − 14n + 45 P n −2 − 8c c 2 − 1 c 2 (n − 4) − n + 8 P n−2 + (n − 4)2 c 2 (n − 1) − n + 9 P n + 4c c 2 n2 − 4n + 3 − n2 + 4n + 29 P n + 4 c 2 − 1 c 2 (n − 1) − n + 9 P n .
(3.16)
Our goal is to eliminate the P n−2 term, then the P n −2 and finally the P n−2 to arrive at a differential equation with just the derivatives of P n in it. We first have to lower the degree of c in the polynomial in front of P n−2 . Thus if we multiply (3.16) by −2c (n − 6) and add it to (3.15) we get
0 = (n − 11)(n − 2)2 c 2 (n + 1) − n + 7 P n−2
+ 4c (n − 11) c 2 n2 − 1 − n2 + 33 P n −2 + 4(n − 11) c 2 − 1 c 2 (n + 1) − n + 7 P n−2 − 8c (n − 11)(n − 4)2 c 2 (n − 7) − n − 1 P n − 4(n − 11) c 2 n2 − 8n + 39 − n2 + 8n − 7 P n − 32c (n − 11) c 2 − 1 P n .
If n = 11 we get
0 = (n − 2)2 c 2 (n + 1) − n + 7 P n−2
+ 4c c 2 n2 − 1 − n2 + 33 P n −2 + 4 c 2 − 1 c 2 (n + 1) − n + 7 P n−2 − 8c (n − 4)2 P n − 4 c 2 n2 − 8n + 39 − n2 + 8n − 7 P n − 32c c 2 − 1 P n .
(3.17)
2860
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
We lower the degree of c in the polynomial in front of P n−2 another time by multiplying this last equation by 2c (n − 4) and add it to n + 1 times (3.14) to get
0 = 8c (n − 9)(n − 2)2 P n−2
+ 4(n − 9) c 2 n2 − 4n + 27 − n2 + 4n + 5 P n −2 + 32(n − 9)c c 2 − 1 P n−2 + (n − 9)(n − 4)2 c 2 (n − 7) − n − 1 P n − 4c (n − 9) c 2 n2 − 12n + 35 − n2 + 12n − 3 P n + 4(n − 9) c 2 − 1 c 2 (n − 7) − n − 1 P n .
So that if n = 9, 11 one has
0 = 8c (n − 2)2 P n−2
+ 4 c 2 n2 − 4n + 27 − n2 + 4n + 5 P n −2 + 32c c 2 − 1 P n−2 + (n − 4)2 c 2 (n − 7) − n − 1 P n − 4c c 2 n2 − 12n + 35 − n2 + 12n − 3 P n + 4 c 2 − 1 c 2 (n − 7) − n − 1 P n .
(3.18)
We lower the degree of c in the polynomial in front of P n−2 another time by multiplying this last equation by c (n + 1) and add it to −8 times (3.17) to get
0 = 8(n − 7)(n − 2)2 P n−2
+ 4c (n − 7) c 2 n2 − 4n − 5 − n2 + 4n + 37 P n −2 + 32 c 2 − 1 (n − 7) P n−2 + c (n − 7)(n − 4)2 c 2 (n + 1) − n − 9 P n − 4(n − 7) c 4 n2 − 4n − 5 − c 2 n2 + 4n − 45 + 8(n − 1) P n + 4c c 2 − 1 (n − 7) c 2 (n + 1) − n − 9 P n .
Which if n = 7, 9, 11, then we have
0 = 8(n − 2)2 P n−2
+ 4c c 2 n2 − 4n − 5 − n2 + 4n + 37 P n −2 + 32 c 2 − 1 P n−2 + c (n − 4)2 c 2 (n + 1) − n − 9 P n − 4 c 4 n2 − 4n − 5 − c 2 n2 + 4n − 45 + 8(n − 1) P n + 4c c 2 − 1 c 2 (n + 1) − n − 9 P n .
(3.19)
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
2861
We now want to get rid of the term with P n−2 in it. This is done by multiplying the previous equation by c and add it to −1 times (3.18) we get
2
0 = 4 c 2 − 1 (n − 5)(n + 1) P n −2
2 + c 2 − 1 (n − 4)2 (n + 1) P n 2 − 4c c 2 − 1 (n − 5)(n + 1) P n 3 + 4 c 2 − 1 (n + 1) P n .
Thus as c = ±1 and we are assuming n = −1, 7, 9, 11 then we have
0 = 4(n − 5) P n −2 + (n − 4)2 P n − 4c (n − 5) P n + 4 c 2 − 1 P n .
(3.20)
If we differentiate this with respect to c we get
0 = 4(n − 5) P n−2 + (n − 4)2 P n − 4(n − 5) P n − 4c (n − 5) P n + 8c P n + 4 c 2 − 1 P n
= 4(n − 5) P n−2 + (n − 6)2 P n − 4c (n − 7) P n + 4 c 2 − 1 P n .
(3.21)
Now we work on the coefficient in front of P n −2 to eliminate it. We can multiply the previous equation by −8(c 2 − 1) and add to n − 5 times (3.19) to give us
0 = 8(n − 5)(n − 2)2 P n−2
+ 4c (n − 5) c 2 n2 − 4n − 5 − n2 + 4n + 37 P n −2 + c (n − 5)(n − 4)2 c 2 (n + 1) − n − 9 P n + 4 c 4 −(n − 5)2 (n + 1) + c 2 n3 − 3n2 − 41n + 153 − 6n2 + 24n + 32 P n + 4c c 2 − 1 8(n − 7) + (n − 5) c 2 (1 + n) − 9 − n P n 2 − 32 c 2 − 1 P n .
Now we multiply (3.20) by c (37 + 4n − n2 + c 2 (n − 5)(n + 1)) and add it to −1 times the equation above to give us
0 = −8(n − 5)(n − 2)2 P n−2 + 8c (n − 4)2 (n − 1) P n
2 − 8 c 2 − 1 3n2 − 12n − 16 P n + 192c c 2 − 1 P n + 32 c 2 − 1 P n .
Thus if c = ±1, n = −1, 7, 9, 11 we have
0 = −(n − 5)(n − 2)2 P n−2 + c (n − 4)2 (n − 1) P n
2 − c 2 − 1 3n2 − 12n − 16 P n + 24c c 2 − 1 P n + 4 c 2 − 1 P n .
2862
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
If we differentiate this with respect to c we get
0 = −(n − 5)(n − 2)2 P n −2
+ (n − 4)2 (n − 1) P n + c n3 − 15n2 + 48n + 16 P n + −c 2 3n2 − 12n − 88 + 3n2 − 12n − 40 P n + 40c c 2 − 1 P n 2 ( i v ) + 4 c2 − 1 P n . Now we multiply this by 4 and add it to (n − 2)2 times (3.20) to get the 4th order linear differential equation satisfied by the polynomials P n . We proved Theorem 3.1.1. The polynomials P n = P −4,n satisfy the following differential equation:
+ 160c c 2 − 1 P n − 8 c 2 n2 − 4n − 46 − n2 + 4n + 22 P n − 24c n2 − 4n − 6 P n + (n − 4)2n2 P n = 0.
16 c 2 − 1
2
(i v )
Pn
(3.22)
3.2. Elliptic case 2 From now on we are going to reindex the polynomials P −2,n :
P −2 (c , z) = z 1 − 2cz2 + z4
1
( z4 − 2cz2 + 1)3/2
=
∞
P −2,n (c ) zn .
n =0
This means that now P −2,2 (c ) = 1, P −2,3 (c ) = P −2,1 (c ) = P −2,0 (c ) = 0 and P −2,n (c ) satisfy the following recursion:
(6 + 2k) P k+4 (c ) = 4kc P k+2 (c ) − 2(k − 3) P k (c ).
(3.23)
The first few nonzero polynomials in c are
P −2,2 (c ) = 1,
P −2,6 = 1/5,
P −2,10 = −7 + 32c 2 /105,
P −2,8 = 8c /35,
P −2,12 = 8c −29 + 64c 2 /1155,
so that
P −2 (c , z) = z2 +
1 5
z6 +
8c 35
z8 +
32c 2 − 7 105
z10 +
8c (64c 2 − 29) 1155
z12 + O z14 .
After a very similar lengthy analysis as in the previous section we arrive at the following result: Theorem 3.2.1. The polynomials P n = P −2,n satisfy the following fourth order linear differential equation:
+ 160c c 2 − 1 P n − 8 c 2 n2 − 4n − 42 − n2 + 4n + 18 P n − 24c n2 − 4n − 2 P n + (n − 6)(n − 2)2 (n + 2) P n = 0.
16 c 2 − 1
2
(i v )
Pn
(3.24)
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
2863
4. Associated ultraspherical polynomials After shifting the indices back by 4 we obtain that both families of the polynomials P −4,n (c ) and P −2,n (c ) satisfy the recurrence relation
2kc P k−2 (c ) = (3 + k) P k (c ) + (k − 3) P k−4 (c ).
(4.1)
Note that all odd polynomials are zero. Set k = 2(n + 1) and q s := P 2s . Then we have
4(n + 1)cqn = (2n + 5)qn+1 + (2n − 1)qn−1 ,
(4.2)
where q s = q s (c ). For q s (c ) = P −4,2s (c ) one has
q−2 = P −4,−4 = 1, q1 = P −4,2 = 4c /5,
q−1 = P −4,−2 = 0,
q2 = P −4,4 =
32c − 5 2
35
,
q0 = P −4,0 = 1, q3 = P −4,6 =
16 2 c 8c − 3 . 105
(4.3)
We will show that they are special cases of the associated ultraspherical polynomials. (ν ) For c and ν two complex constants, if one lets C n (x; c ) denote the associated ultraspherical poly(ν ) (ν ) nomials with initial conditions C −1 (x; c ) = 0, C 0 (x; c ) = 1 then they are defined so as to satisfy the difference equation (ν )
(ν )
(ν )
2x(n + ν + c )C n (x; c ) = (n + c + 1)C n+1 (x; c ) + (2ν + n + c − 1)C n−1 (x; c ).
(4.4)
Then setting c = 3/2 and ν = −1/2 this equation is the same recurrence relation as (4.2). Hence, we can conclude that polynomials P −4,n is a special case of associated ultraspherical polynomials. (α ,β)
For α , β, c ∈ C constants, the associated Jacobi polynomials P n the recurrence relation
(x; c ) are defined so as to satisfy
2(n + c + 1)(n + c + γ )(2n + 2c + γ − 1) pn+1
= (2n + 2c + γ ) (2n + 2c + γ − 1)(2n + 2c + γ + 1)x + (γ − 1)(γ − 2β − 1) pn − 2(n + c + γ − β − 1)(n + c + β)(2n + 2c + γ + 1) pn−1 , (α ,β)
(4.5)
(α ,β)
where n ∈ N, γ = α + β + 1, P −1 (x, c ) = 0; P 0 (x, c ) = 1. One can check that the associated ultraspherical polynomials are related to the associated Jacobi polynomials through the formula (ν −1/2,ν −1/2)
Pn
(x; β) =
(ν + c + 12 )n (ν ) C n (x; β). (2ν + c )n
The weight measure for the associated ultraspherical polynomials is known (see [2]) and is given by
ρ (cos t ) =
(2 sin t )2β−1 [Γ (λ + γ )]2 2ı t −2 2 F 1 1 − β, γ ; γ + β; e 2π Γ (2β + γ )Γ (γ + 1)
(4.6)
2864
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
for 0 t π and where 2 F 1 (a, b; c ; z) is the hypergeometric function 2 F 1 (a, b ; c ; z ) =
∞ (a)n (b)n n =0
(c )n
zn .
Corollary 4.0.2. Polynomials P −4,n are non-classical orthogonal polynomials. Finally we should point out that Wimp (see [13], Eqs. 5.7.10–5.7.12 and formula (48) in [30]) using MACSYMA and ideas from [29], deduced that the associated Jacobi Polynomials satisfy the fourth order differential equation
A 0 (x) y ( vi ) + A 1 (x) y + A 2 (x) y + A 3 (x) y + A 4 (x) y = 0,
(4.7)
where
A 0 (x) = 1 − x2
2
,
A 1 (x) = 10x x2 − 1 ,
A 2 (x) = −(1 − x)2 2K + 2C + γ 2 − 25 + 2(1 − x)(2K + 2C + 2αγ ) + 2(α + 1) − 26,
A 3 (x) = 3(1 − x) 2K + 2C + γ 2 − 5 − 6( K + C + αγ + β − 2), A 4 (x) = n(n + 2)(n + γ + 2c )(n + γ + 2c − 2), where
K = (n + c )(n + γ + c ),
C = (c − 1)(c + α + β). (−1/2)
In the setting of the polynomials qn (x) = P −4,2n+4 (x) = C n differential equation we arrived at takes the form
2
(x, 3/2) for n −1, the fourth order
+ 10x x2 − 1 qn − x2 2n2 + 4n − 23 − 2n2 − 4n + 11 qn − 3 · x 2n2 + 4n − 3 qn + n2 (n + 2)2 qn = 0. (i v )
x2 − 1 qn
For n = 2 we get q2 (x) =
32x2 −5 , 35
and plugging this into the equation above we have
− −7x2 − 5 64 − 3 · x · 13 · 64x + 4 · 16 32x2 − 5 = 0. In Ismail’s notation we have c = 3/2,
ν = −1/2, α = β = γ = −1,
K = (n + 3/2)(n + 1/2),
C = −1/4
and
A 2 (x) = − 2n2 + 4n − 23 x2 − 52x + 2n2 + 4n + 3,
A 3 (x) = −3 2n2 + 4n − 3 x, A 4 (x) = n2 (n + 2)2 .
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
2865
When n = 2 we see that A 2 (x) is slightly off from our corresponding coefficient and q2 (x) = 32x35−5 does not satisfy the above equation. So there seems to be a typo in the formulation of A 2 (x) given in [13] and the families of fourth order differential equations (in the variable x) given in the book by Ismail for the associated ultraspherical polynomials are different from the families of fourth order linear differential equations (in the variable c) derived in our work above, i.e. (3.22) and (3.24). 2
5. Orthogonality of P −2,n Polynomials P −2,n satisfy the same recurrence relation as polynomials P −4,n but have different initial conditions. Hence, we do know immediately whether they are associated ultraspherical polynomials. Here we will give an independent proof of the orthogonality of these polynomials. For q s (c ) = P −2,2s (c ) we have
q−1 = P −2,−2 = 1,
q0 = P −2,0 = 0,
q2 = P −2,4 = 8c /35,
q1 = P −2,2 = 1/5,
q3 = P −2,6 =
32c 2 − 7 105
.
If q s (c ) = P −2,2s (c ) we want polynomials with index n giving the degree of the polynomial, then we will set
q¯ n := qn+1 ,
n −1,
while we ignore the “first” two polynomials; q−1 = 1 and q0 = 0. Then
q¯ 0 = 1/5,
q¯ 1 = 8c /35,
q¯ 2 =
32c 2 − 7 105
,
q¯ 3 =
8c (64c 2 − 29) 1155
(5.1)
and (4.2), becomes
4(n + 2)cqn+1 = (2n + 7)qn+2 + (2n + 1)qn ,
(5.2)
4(n + 2)cq¯ n = (2n + 7)¯qn+1 + (2n + 1)¯qn−1 ,
(5.3)
or
where the polynomials q¯ n are of degree n in c. We recall the following result by Favard which can be found in [8] or in Theorem 4.4 of [5]: Theorem 5.0.3. Let {φn , n 0} be a sequence of polynomials (monic) where pn is a polynomial of degree n, satisfying the following recursion
φn = (x − μn )φn−1 − λn φn−2 ,
n = 1, 2, 3, . . . ,
(5.4)
for some complex numbers μn , λn , n = 1, 2, 3, . . . and φ−1 = 0, φ0 = 1. Then {φn , n 0} is a sequence of orthogonal polynomials with respect to a (unique) weight measure if and only if μn ∈ R and λn+1 > 0 for all n 1. We have the following equivalent version of the theorem above.
2866
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
Theorem 5.0.4. Let { pn , n 0} be a sequence of polynomials where pn is a polynomial of degree n, satisfying the following recursion
xpn = an+1 pn+1 + bn pn + cn−1 pn−1 ,
n = 0, 1, 2, . . . ,
(5.5)
for some complex numbers an+1 , bn , cn−1 , n = 0, 1, 2, . . . and p −1 = 0. Then { pn , n 0} is a sequence of orthonormal polynomials with respect to some (unique) weight measure if and only if bn ∈ R and cn = a¯ n+1 = 0 for all n 0. Proof. The equivalence of this statement with the original result of Favard is established in the following way: Let kn = ||φn || and put pn = kn−1 φn . Then from (5.4) we get
xpn = kn−1 kn+1 pn+1 + μn+1 pn + kn−1 λn+1 kn−1 pn−1 . By putting an+1 = kn−1 kn+1 , bn = μn+1 and cn−1 = kn−1 λn+1 kn−1 we obtain (5.5) with bn ∈ R and cn = a¯ n+1 = 0 for all n 0, because
cn = (xpn+1 , pn ) = ( pn+1 , xpn ) = a¯ n+1 . Conversely, let dn be the leading coefficient of pn and put φn = dn−1 pn . Then from (5.5) we get
xφn−1 = dn−−11 an dn φn + bn−1 φn−1 + dn−−11 cn−2 dn−2 φn−2 . Since the φn ’s are monic we have dn−−11 an dn = 1. Therefore we obtain (5.4) with λn+1 = dn−1 cn−1 dn−1 = an cn−1 = an a¯ n > 0 for all n 1. 2
μn = bn−1 ∈ R and
+7 2n+1 We apply Theorem 5.0.4 to our setting. If we set an+1 = 42n (n+2) , bn = 0, cn−1 = 4(n+2) , then the hypothesis of Favard’s Theorem is not satisfied. But with a modification to our recursion formula we show that the q¯ n are orthogonal.
Theorem 5.0.5. The polynomials P −2,n (c ) are orthogonal with respect to some weight function. Proof. It is sufficient to check that there exist a family of orthonormal polynomials f n and constants λn such that qn = λn f n for all n We have the following recursion for f n :
4(n + 2)c f n = (2n + 7)λn+1 f n+1 + (2n + 1)λn−1 f n−1 or
c fn =
(2n + 7)λn+1 (2n + 1)λn−1 f n +1 + f n −1 , 4(n + 2)λn 4(n + 2)λn
for n 1. Set
A n +1 =
(2n + 7)λn+1 , 4(n + 2)λn
C n −1 =
(2n + 1)λn−1 . 4(n + 2)λn
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
(2n+5)λ
Then A n = C n−1 if and only if 4(n+1)λ n = n −1
λn2 =
2867
(2n+1)λn−1 or 4(n+2)λn
(n + 1)(2n + 1) 2 λ . (n + 2)(2n + 5) n−1
Taking λ0 = 1 we can find a family of constants λn satisfying this relation. Hence, by Theorem 5.0.4 the polynomials f n form an orthonormal family with respect to some measure, and therefore the polynomials are orthogonal. 2 One can also show that P −2,n are non-classical orthogonal polynomials. We will give a simple proof of this fact in Appendix A. The same argument works in the case of P −4,n . Acknowledgments The second author was supported in part by the CNPq grant (301743/2007-0), by the Fapesp grant (2010/50347-9) and by the MathAmSud grant (611/2012). Appendix A Given a sequence of orthogonal polynomials {qn }n0 we consider the complex linear space of all differential operators with complex coefficients on the real line that have the polynomials qn as their eigenfunctions. Thus
D( w ) = D: Dqn = γn ( D )qn , γn ( D ) ∈ C for all n 0 . Some properties of D ( w ), see [12]: (1) The definition of D ( w ) depends only on the weight function w = w (x) and not on the orthogonal sequence {qn }n0 . (2) If D ∈ D ( w ), then
D=
s
i d f i (x) , dx
i =0
where f i (x) is a polynomial and deg f i i. To ease the notation if ν ∈ C let
[ν ]i = ν (ν − 1) · · · (ν − i + 1), s
[ν ]0 = 1.
(3) If {qn }n0 is a sequence of orthogonal polynomials and D ∈ D ( w ) is of the form D =
i =0
d i f i (x)( dx ) , with f i (x) =
i
j =0
f ji ( D )x j , then
γn ( D ) =
s [n]i f ii ( D ). i =0
Let us consider the Weyl algebra D = { D =
D=
D=
s i =0
s
i =0
d i f i (x)( dx ) : f i ∈ C[x]}, and the subalgebra
i d f i (x) ∈ D: deg( f i ) i . dx
2868
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
(4) If D ∈ D satisfies the symmetry condition ( Dp , q) = ( p , Dq) for all p , q ∈ C[x], then D ∈ D ( w ). (5) For any D ∈ D ( w ) there is a unique D ∗ ∈ D ( w ) such that ( Dp , q) = ( p , D ∗ q) for all p , q ∈ C[x]. We refer to D ∗ as the adjoint of D. The map D → D ∗ is a ∗-operation in the algebra D ( w ), and the orders of D and D ∗ coincide. (6) The set S ( w ) = { D ∈ D ( w ): D = D ∗ } of all symmetric differential operators is a real form of D ( w ):
D( w ) = S ( w ) ⊕ i S ( w ). (7) D ∈ D ( w ) is symmetric if and only if γn ( D ) is real for all n 0. In the literature a weight w is called classical if there exists a second order symmetric differential operator D such that Dqn = γn qn for all n 0. Theorem A.0.6. The sequences of orthogonal polynomials (5.1) {¯qn }n0 and (4.3) {qn }n0 are not classical. Proof. In the polynomials above we replace c by x as it is more natural to use this later letter as our variable. From (6) above, to prove the theorem is equivalent to prove that in D ( w ) there is no differential operator of order 2. Suppose D ∈ D ( w ) is of order 2. Then we can assume that D is of the form
D = ax2 + bx + c
2
d
+ (ex + f )
dx
d dx
(A.1)
,
for some a, b, c , e , f ∈ C. For (5.1) the first five terms of the orthogonal sequence {¯qn }n0 are:
q¯ 0 =
1 5
q¯ 1 =
,
q¯ 4 =
8x 35
q¯ 2 =
,
32x2 − 7 105
,
q¯ 3 =
8x(64x2 − 29) 1155
,
(160 × 64)x4 − (32 × 222)x2 + (77 × 7) . 13 × 1155
From the definition of D ( w ) and (3) we know that
ax2 + bx + c
2
d dx
q¯ n + (ex + f )
d dx
q¯ n = an(n − 1) + bn + c + en + f q¯ n ,
(A.2)
for all n 0. If we set n = 0 we get c + f = 0. If we put n = 1 in the above equation we get
8
(ex + f )
35
= (b + c + e + f )
8x 35
,
which implies f = 0 and so c = 0 (as c + f = 0) and e = b + e. Hence b = 0. Then for n = 2 we obtain
(2a + 2e )
32x2 − 7 105
2
= ax
= ax2
d
2
32x2 − 7
dx 64 105
105
+ ex
+ ex
d dx
32x2 − 7
105
64x 105
and hence a + e = 0. Using all this information, the above Eq. (A.2) for n = 3 is equivalent to
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
(6a + 3e )
8x(64x2 − 29) 1155
= ax2 =
d
2
8x(64x2 − 29)
dx
1155
8 · 64 · 6 · ax3 1155
+
+ ex
8 · 64 · 3 · ex3 1155
−
d
2869
8x(64x2 − 29)
dx
8 · 29ex 1155
1155
,
which reduces to
2a + e = 0. Therefore a = e = 0 which implies that D = 0. In other words we have proved that the only differential operators in D ( w ) of degree less or equal to two are the constants. For (4.3)
q 0 = 1,
q1 = 4x/5,
A similar analysis yields also D = 0.
q2 =
32x2 − 5 35
,
q3 =
16 2 x 8x − 3 .
105
2
References [1] George B. Arfken, Hans J. Weber, Mathematical Methods for Physicists, fifth edition, Harcourt/Academic Press, Burlington, MA, 2001. [2] Joaquin Bustoz, Mourad E.H. Ismail, The associated ultraspherical polynomials and their q-analogues, Canad. J. Math. 34 (3) (1982) 718–736. [3] Murray Bremner, Universal central extensions of elliptic affine Lie algebras, J. Math. Phys. 35 (12) (1994) 6685–6692. [4] Ben Cox, Vyacheslav Futorny, DJKM algebras I: their universal central extension, Proc. Amer. Math. Soc. 139 (10) (2011) 3451–3460. [5] T.S. Chihara, An Introduction to Orthogonal Polynomials, Dover Publishers, New York, 2011. [6] Etsuro¯ Date, Michio Jimbo, Masaki Kashiwara, Tetsuji Miwa, Landau–Lifshitz equation: solitons, quasiperiodic solutions and infinite-dimensional Lie algebras, J. Phys. A 16 (2) (1983) 221–236. [7] Pavel I. Etingof, Igor B. Frenkel, Alexander A. Kirillov Jr., Lectures on Representation Theory and Knizhnik–Zamolodchikov Equations, Math. Surveys Monogr., vol. 58, Amer. Math. Soc., Providence, RI, 1998. [8] J. Favard, Sur les polynomes de Tchebicheff, C. R. Acad. Sci. Paris 200 (1935) 2052–2053. [9] Boris Feigin, Edward Frenkel, Integrable hierarchies and Wakimoto modules, in: Differential Topology, Infinite-Dimensional Lie Algebras, and Applications, in: Amer. Math. Soc. Transl. Ser. 2, vol. 194, Amer. Math. Soc., Providence, RI, 1999, pp. 27–60. [10] Edward Frenkel, Wakimoto modules, opers and the center at the critical level, Adv. Math. 195 (2) (2005) 297–404. [11] Edward Frenkel, Langlands Correspondence for Loop Groups, Cambridge Stud. Adv. Math., vol. 103, Cambridge University Press, Cambridge, 2007. [12] F. Alberto Grünbaum, Juan Tirao, The algebra of differential operators associated to a weight matrix, Integral Equations Operator Theory 58 (4) (2007) 449–475. [13] Mourad E.H. Ismail, Classical and Quantum Orthogonal Polynomials in One Variable, Encyclopedia Math. Appl., vol. 98, Cambridge University Press, Cambridge, 2005, with two chapters by Walter Van Assche, with a foreword by Richard A. Askey. [14] Igor Moiseevich Krichever, S.P. Novikov, Algebras of Virasoro type, Riemann surfaces and strings in Minkowski space, Funktsional. Anal. i Prilozhen. 21 (4) (1987) 47–61, 96. [15] Igor Moiseevich Krichever, S.P. Novikov, Algebras of Virasoro type, Riemann surfaces and the structures of soliton theory, Funktsional. Anal. i Prilozhen. 21 (2) (1987) 46–63. [16] Igor Moiseevich Krichever, S.P. Novikov, Algebras of Virasoro type, the energy-momentum tensor, and operator expansions on Riemann surfaces, Funktsional. Anal. i Prilozhen. 23 (1) (1989) 24–40. [17] H.L. Krall, On Orthogonal Polynomials Satisfying a Certain Fourth Order Differential Equation, Pennsylvania State College Studies, vol. 6, 1940, p. 24. [18] Allan M. Krall, Orthogonal polynomials satisfying fourth order differential equations, Proc. Roy. Soc. Edinburgh Sect. A 87 (3–4) (1980/1981) 271–288. [19] M. Kaneko, D. Zagier, Supersingular j-invariants, hypergeometric series, and Atkin’s orthogonal polynomials, in: Computational Perspectives on Number Theory, Chicago, IL, 1995, in: AMS/IP Stud. Adv. Math., vol. 7, Amer. Math. Soc., Providence, RI, 1998, pp. 97–126. [20] Lance L. Littlejohn, Samuel D. Shore, Nonclassical orthogonal polynomials as solutions to second order differential equations, Canad. Math. Bull. 25 (3) (1982) 291–295.
2870
B. Cox et al. / J. Differential Equations 255 (2013) 2846–2870
[21] Martin Schlichenmaier, Higher genus affine algebras of Krichever–Novikov type, Mosc. Math. J. 3 (4) (2003) 1395–1427. [22] Martin Schlichenmaier, Local cocycles and central extensions for multipoint algebras of Krichever–Novikov type, J. Reine Angew. Math. 559 (2003) 53–94. g ,2 and sl g ,2 , in: Fundamental Mathe[23] O.K. Sheinman, Second-order Casimirs for the affine Krichever–Novikov algebras gl matics Today, Nezavis. Mosk. Univ., Moscow, 2003, pp. 372–404 (in Russian). [24] O.K. Sheinman, Highest-weight representations of Krichever–Novikov algebras and integrable systems, Uspekhi Mat. Nauk 60 (2(362)) (2005) 177–178. [25] M. Schlichenmaier, O.K. Scheinman, The Sugawara construction and Casimir operators for Krichever–Novikov algebras, in: Complex Analysis and Representation Theory, vol. 1, J. Math. Sci. (N. Y.) 92 (2) (1998) 3807–3834. [26] M. Shlikhenmaier, O.K. Sheinman, The Wess–Zumino–Witten–Novikov theory, Knizhnik–Zamolodchikov equations, and Krichever–Novikov algebras, Uspekhi Mat. Nauk 54 (1(325)) (1999) 213–250. [27] V.V. Schechtman, A.N. Varchenko, Hypergeometric solutions of Knizhnik–Zamolodchikov equations, Lett. Math. Phys. 20 (4) (1990) 279–283. (1) [28] Minoru Wakimoto, Fock representations of the affine Lie algebra A 1 , Comm. Math. Phys. 104 (4) (1986) 605–609. [29] G.N. Watson, A Treatise on the Theory of Bessel Functions, Cambridge Math. Lib., Cambridge University Press, Cambridge, 1995, reprint of the second (1944) edition. [30] Jet Wimp, Explicit formulas for the associated Jacobi polynomials and some applications, Canad. J. Math. 39 (4) (1987) 983–1000.